Abstract Algebra — UPSC Maths Optional Notes

1. Understanding Groups and the Group Axioms

Why we need it: integers under addition, non-zero reals under multiplication, the rotations of a square and the shuffles of a deck of cards all obey the same few rules.

Instead of proving a fact such as 'inverses are unique' separately for each of them, we prove it once for every structure that obeys these rules. A group is exactly this list of rules.

In words: 'a non-empty set G' means G has at least one element (it will contain the identity). 'A binary operation *' is a rule that takes an ordered pair (a, b) of elements of G and returns one element a*b.

Closure says the answer a*b lies in G again. Associativity says that when three elements are combined, the bracketing does not matter; the order of the elements may still matter.

The identity e is one fixed element that changes nothing, on either side. An inverse of a is an element that undoes a, on either side.

Notice the order of the quantifiers. There is one e that works for every a (∃e ∀a), but the inverse depends on a (∀a ∃a⁻¹).

First example, slowly: let G = {1, −1, i, −i} with ordinary multiplication of complex numbers. We check the four axioms one at a time.

Step 1 (closure): i·i = −1, i·(−i) = 1, (−i)·(−i) = −1, (−1)·i = −i, (−1)·(−1) = 1, and every other product is ±1 or ±i. So every product lies in G.

Step 2 (associativity): multiplication is associative in all of ℂ, and G ⊆ ℂ, so it is associative on G. Nothing needs to be checked by hand.

Step 3 (identity): 1·z = z·1 = z for each z ∈ G, so e = 1.

Step 4 (inverses): 1·1 = 1, (−1)(−1) = 1 and i·(−i) = 1. So 1⁻¹ = 1, (−1)⁻¹ = −1, i⁻¹ = −i and (−i)⁻¹ = i, all in G.

Step 5 (conclusion): G is a group, and it is abelian because complex multiplication commutes. Its order is 4, and o(i) = 4 because i, i² = −1, i³ = −i, i⁴ = 1.

Non-example: {0, 1, 2, 3} under multiplication mod 4 is closed and has the identity 1, but 2 has no inverse: 2·0 = 0, 2·1 = 2, 2·2 = 0, 2·3 = 2 (mod 4), never 1. So axiom (iv) fails.

Removing 0 does not help, because then 2·2 = 0 breaks closure. Only the units {1, 3} = U(4) form a group under multiplication mod 4.

Building up: the basic facts follow from the axioms alone, one axiom per step. (1) The identity is unique: if e and e′ are both identities, then e = ee′ = e′.

(2) Inverses are unique: if b and c both invert a, then b = be = b(ac) = (ba)c = ec = c. Here associativity is used in the middle step.

(3) Cancellation: if ab = ac, multiply on the left by a⁻¹ and use associativity to get b = c. (4) (ab)⁻¹ = b⁻¹a⁻¹, because (ab)(b⁻¹a⁻¹) = a(bb⁻¹)a⁻¹ = aa⁻¹ = e.

In an exam answer, name the axiom used in each such step; that is what the examiner looks for.

Where it is used: every later topic (subgroups, homomorphisms, rings, vector spaces) begins by checking these axioms, and uniqueness and cancellation are used silently in nearly every proof of this chapter.

2. Understanding Subgroups and the One-Step Test

Why we need it: a big group often contains smaller groups: the even integers inside ℤ, the rotations inside all the symmetries of a square. We want a quick test for when a subset is itself a group, without re-checking all four axioms.

In words: H ≤ G means H ⊆ G and H is a group under the same operation as G. The one-step test asks two things: H is not empty, and for every a, b ∈ H the element ab⁻¹ is again in H.

Here b⁻¹ is the inverse in G. In additive notation the condition reads a − b ∈ H: 'the difference of two members is a member'.

First example, slowly: let H = 3ℤ = {..., −6, −3, 0, 3, 6, ...} in (ℤ, +).

Step 1: H is not empty, since 0 = 3·0 ∈ H.

Step 2: take arbitrary a, b ∈ H. Membership means a = 3m and b = 3n for some integers m, n.

Step 3: in additive notation ab⁻¹ is a − b = 3m − 3n = 3(m − n).

Step 4: m − n is an integer, so a − b ∈ H, and by the one-step test 3ℤ ≤ ℤ. Notice that we never checked associativity (it is inherited from ℤ) or the identity (the test produces it).

Non-example: the odd integers in (ℤ, +) form a non-empty set, but 3 − 1 = 2 is even, so the test fails; indeed 0 is not odd. Also ℕ = {1, 2, 3, ...} is closed under +, yet 1 − 2 = −1 ∉ ℕ: for an infinite subset, closure alone is not enough.

Building up: why one condition is enough. (i) Put b = a: aa⁻¹ = e ∈ H. (ii) Put a = e: eb⁻¹ = b⁻¹ ∈ H, so H contains inverses.

(iii) For a, b ∈ H we now know b⁻¹ ∈ H, so a(b⁻¹)⁻¹ = ab ∈ H: H is closed. (iv) Associativity holds in G, hence in H.

For a finite H, closure alone gives the inverses as well, because a⁻¹ = aⁿ⁻¹ with n = o(a); this is the finite subgroup test stated below.

Where it is used: to prove that kernels, centres, centralisers, intersections and images are subgroups; the same idea is the subspace test of Linear Algebra and the subring and ideal tests of the Rings topic.

3. Understanding Cyclic Groups and Orders of Elements

Why we need it: the simplest groups are those built from a single element and its powers. They are completely understood, so they supply ready examples, and the rule o(a^k) = n/gcd(n, k) answers most counting questions.

In words: ⟨a⟩ = {aⁿ : n ∈ ℤ} is the set of all powers of a, including a⁰ = e and the negative powers a⁻ⁿ = (a⁻¹)ⁿ. G is cyclic if G = ⟨a⟩ for some a, called a generator.

The order o(a) is the least positive n with aⁿ = e. When it is finite, it equals the number of elements of ⟨a⟩. In additive notation 'aⁿ' means na = a + a + ... + a (n terms).

First example, slowly: work in ℤ₁₂ under addition. The 'powers' of 8 are its multiples: 1·8 = 8, 2·8 = 16 ≡ 4, 3·8 = 24 ≡ 0.

So o(8) = 3 and ⟨8⟩ = {0, 4, 8}. The formula agrees: 12/gcd(12, 8) = 12/4 = 3.

Now take 5: its multiples are 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 0, all twelve residues. So 5 generates ℤ₁₂, matching gcd(5, 12) = 1.

What to notice: the size of ⟨a⟩ depends only on how many steps a takes to return to the identity.

Non-example: U(8) = {1, 3, 5, 7} under multiplication mod 8 is abelian of order 4, but it is not cyclic: 3² ≡ 5² ≡ 7² ≡ 1, so each ⟨a⟩ has at most 2 elements. Abelian does not imply cyclic.

Building up: (1) If o(a) = n, the elements e, a, ..., aⁿ⁻¹ are distinct, and a^m depends only on the remainder of m mod n; so |⟨a⟩| = n.

(2) a^m = e iff n divides m, by the division algorithm (the 'Why it works' line below). (3) Hence o(a^k) = n/gcd(n, k).

(4) a^k generates exactly when this order is n, i.e. gcd(n, k) = 1, giving φ(n) generators (Standard Proofs, Theorem 16). (5) Every subgroup of ⟨a⟩ is cyclic (Theorem 2), and there is exactly one for each divisor of n.

Where it is used: listing all subgroups of ℤ_n and U(n), counting elements of a given order, the cyclic multiplicative group of a finite field, and the structure of groups of prime order.

4. Understanding Cosets and Lagrange's Theorem

Why we need it: we want to know which sizes a subgroup of a finite group can have, without listing subgroups. Cosets cut the group into equal pieces, and counting the pieces answers the question.

In words: for H ≤ G and a ∈ G, the left coset aH = {ah : h ∈ H} is obtained by multiplying every element of H on the left by a. The element a is a representative of the coset.

Different representatives can give the same coset: aH = bH exactly when a⁻¹b ∈ H. The index [G : H] is the number of distinct left cosets.

First example, slowly: take G = ℤ₆ under addition and H = {0, 3}. In additive notation the coset of a is a + H.

Step 1: 0 + H = {0, 3}, 1 + H = {1, 4}, 2 + H = {2, 5}.

Step 2: 3 + H = {3, 0} = H again, 4 + H = {4, 1} = 1 + H and 5 + H = 2 + H. So there are exactly three distinct cosets, and [G : H] = 3.

Step 3: the three cosets have 2 elements each, do not overlap, and together contain all six elements. Hence 6 = 3 × 2, i.e. |G| = [G : H]·|H|.

Non-example: translate the subset S = {0, 1}, which is not a subgroup: 0 + S = {0, 1} and 1 + S = {1, 2} overlap in 1 without being equal. Cosets of a genuine subgroup are always equal or disjoint, and that is what the counting needs.

Building up: (1) The relation a ~ b iff a⁻¹b ∈ H is an equivalence relation, and its classes are the left cosets, so the cosets partition G.

(2) h ↦ ah is a bijection H → aH, so each coset has |H| elements. (3) Counting gives |G| = [G : H]·|H|, which is Lagrange's theorem; the full proof is Standard Proofs, Theorem 1.

(4) Apply it to H = ⟨a⟩: o(a) divides o(G), so a^o(G) = e. (5) If o(G) = p is prime, any a ≠ e has o(a) = p, so G = ⟨a⟩ is cyclic.

Where it is used: Fermat's and Euler's theorems, groups of prime order, the count of homomorphisms, index-2 subgroups being normal, and all Sylow-type counting in later topics.

5. Group and Basic Properties

Group: a non-empty set G with a binary operation * such that (i) closure: a*b ∈ G, (ii) associativity: (a*b)*c = a*(b*c), (iii) an identity e with a*e = e*a = a, (iv) every a has an inverse a⁻¹ with a*a⁻¹ = a⁻¹*a = e.

Idea: a group is the algebra of reversible actions, such as the symmetries of a figure or the shuffles of a deck. Closure keeps us inside the set, associativity lets us drop brackets, the identity is the action 'do nothing' and the inverse is 'undo'.

Why each axiom is needed: (ℤ, ·) has closure, associativity and identity 1 but 2 has no inverse; (ℤ, −) is closed but not associative, since (1 − 2) − 3 = −4 while 1 − (2 − 3) = 2; (ℕ, +) has no identity and no inverses.

Abelian group: a group in which a*b = b*a for all a, b ∈ G.

Basic facts: the identity and each inverse are unique; (ab)⁻¹ = b⁻¹a⁻¹; left and right cancellation laws hold.

Order: o(G) = |G| is the number of elements; o(a) is the least n ≥ 1 with aⁿ = e, and o(a) = ∞ if no such n exists.

Idea: o(a) is the length of the cycle e, a, a², ... before it returns to e. In a finite group the powers must repeat, so aⁱ = a^j with i > j gives a^i−j = e; hence every element of a finite group has finite order. In (ℤ, +) every non-zero element has infinite order, while in ℂ^* the element i has order 4.

Why it works (a^m = e ⇒ o(a) | m): write m = qn + r with n = o(a) and 0 ≤ r < n. Then e = a^m = (aⁿ)^qa^r = a^r, and minimality of n forces r = 0.

Results on order: if a^m = e then o(a) divides m; o(a) = o(a⁻¹) = o(gag⁻¹); o(ab) = o(ba).

Abelian criteria: G is abelian if (ab)² = a²b² for all a, b, or if a² = e for every a ∈ G.

6. Subgroups

Subgroup: a non-empty subset H ⊆ G that is itself a group under the operation of G; written H ≤ G.

One-step test: a non-empty subset H is a subgroup iff a, b ∈ H ⇒ ab⁻¹ ∈ H.

Idea: the single condition ab⁻¹ ∈ H packs together all the subgroup axioms. Taking b = a gives e ∈ H; then a = e gives b⁻¹ ∈ H; then a(b⁻¹)⁻¹ = ab ∈ H. Associativity is inherited from G. Non-emptiness is essential, because the empty set satisfies the implication vacuously.

Finite subgroup test: a non-empty finite subset H is a subgroup iff it is closed under the operation.

Why it works: for a ∈ H, closure puts a, a², a³, ... in H; since H is finite, o(a) = n is finite, so e = aⁿ ∈ H and a⁻¹ = aⁿ⁻¹ ∈ H. Finiteness is essential: ℕ is closed under + in (ℤ, +) but is not a subgroup.

Union and product: the intersection of subgroups is a subgroup; H ∪ K is a subgroup iff H ⊆ K or K ⊆ H; HK is a subgroup iff HK = KH.

Idea: a union of two subgroups is a subgroup only in the trivial case. If h ∈ H∖K and k ∈ K∖H, then hk ∈ H would give k = h⁻¹(hk) ∈ H, and hk ∈ K would give h ∈ K; so hk lies outside H ∪ K. For example 2ℤ ∪ 3ℤ contains 2 and 3 but not 5.

Product formula: for finite subgroups H and K, |HK| = |H|·|K|/|H ∩ K|.

Why it works: consider the map H × K → HK, (h, k) ↦ hk. Now hk = h′k′ iff h⁻¹h′ = k(k′)⁻¹, and this common element lies in H ∩ K. So each element of HK is hit exactly |H ∩ K| times, and |H||K| = |HK|·|H ∩ K|.

Warning example: in S₃, H = {e, (1 2)} and K = {e, (1 3)} give |HK| = 2·2/1 = 4, which does not divide 6; so HK is a subset of S₃ but not a subgroup.

Centre and centraliser: Z(G) = {z ∈ G : zg = gz for all g ∈ G} and C(a) = {g ∈ G : ga = ag} are subgroups of G.

7. Cyclic Groups

Cyclic group: G = ⟨a⟩ = {aⁿ : n ∈ ℤ} for some element a, called a generator.

Idea: a cyclic group is one generator and all its powers: a clock face (ℤ_n) or the integer line (ℤ). Everything about it, its subgroups, generators and quotients, is decided by the order of the generator.

Structure theorem: an infinite cyclic group is isomorphic to (ℤ, +); a cyclic group of order n is isomorphic to (ℤ_n, +_n).

Theorem: every cyclic group is abelian, and every subgroup of a cyclic group is cyclic.

Order of a power: if o(a) = n then o(a^k) = n/gcd(n, k).

Why it works: let d = gcd(n, k). Then (a^k)^m = e ⇔ n | km ⇔ (n/d) | (k/d)m ⇔ (n/d) | m, the last step because gcd(n/d, k/d) = 1. So the least such m is n/d. Check in ℤ₁₂: 8 has order 12/gcd(12, 8) = 3, and indeed 8, 16 ≡ 4, 24 ≡ 0.

Generators: if o(a) = n, then a^k generates ⟨a⟩ iff gcd(k, n) = 1, so there are φ(n) generators. An infinite cyclic group has exactly two generators, a and a⁻¹.

Groups of units: U(n) = {k : 1 ≤ k ≤ n, gcd(k, n) = 1} under multiplication mod n has order φ(n). U(p) is cyclic for every prime p (a generator is called a primitive root mod p); more generally U(n) is cyclic iff n = 1, 2, 4, p^k or 2p^k with p an odd prime.

Infinite groups: (ℤ, +) has the distinct subgroups nℤ (n = 0, 1, 2, ...). (ℚ, +) is not cyclic, and its cyclic subgroups ⟨1/n⟩ = {k/n : k ∈ ℤ} are distinct for different n ≥ 1.

Subgroups of a finite cyclic group: for each divisor d of n there is exactly one subgroup of order d, namely ⟨a^n/d⟩.

Idea: see the figure above: the subgroup lattice of ℤ₁₂ is a copy of the divisor lattice of 12. In general ⟨a^n/d⟩ ⊆ ⟨a^n/e⟩ iff d | e, so containment of subgroups mirrors divisibility of their orders.

Example 1: Find all generators and all subgroups of ℤ₁₂.

Generators are k with gcd(k, 12) = 1: 1, 5, 7, 11, so φ(12) = 4. There is one subgroup for each divisor 1, 2, 3, 4, 6, 12: ⟨0⟩, ⟨6⟩ = {0, 6}, ⟨4⟩ = {0, 4, 8}, ⟨3⟩ = {0, 3, 6, 9}, ⟨2⟩ of order 6, and ℤ₁₂.

8. Cosets and Lagrange's Theorem

Coset: for H ≤ G and a ∈ G, the left coset is aH = {ah : h ∈ H} and the right coset is Ha = {ha : h ∈ H}.

Idea: a coset is a translate of H. In (ℝ², +) with H a line through the origin, the cosets are the lines parallel to H. In ℤ₁₂ with H = ⟨4⟩ = {0, 4, 8}, the cosets are {0, 4, 8}, {1, 5, 9}, {2, 6, 10}, {3, 7, 11} (see the figure above): four blocks of equal size.

Properties: aH = bH iff a⁻¹b ∈ H; two left cosets are either equal or disjoint; |aH| = |H|; the left cosets partition G.

Index: [G : H] is the number of distinct left cosets, which equals the number of distinct right cosets.

Tower law for index: if K ≤ H ≤ G and G is finite, then [G : K] = [G : H][H : K], because o(G)/o(K) = (o(G)/o(H))·(o(H)/o(K)). It holds in any group whenever the indices are finite.

Lagrange's theorem: if G is a finite group and H ≤ G, then o(H) divides o(G), and [G : H] = o(G)/o(H).

Idea: the left cosets cut G into [G : H] disjoint blocks, each of size |H|, so |G| = [G : H]·|H|; the full proof is in Standard Proofs, Theorem 1. Finiteness is needed only for the divisibility statement; in an infinite group the index can still be finite, for example [ℤ : nℤ] = n.

Corollaries: in a finite group, o(a) divides o(G) and a^o(G) = e; every group of prime order p is cyclic and is generated by each of its non-identity elements.

Number theory: applying Lagrange to the group U(n) of units mod n gives Euler's theorem a^φ(n) ≡ 1 (mod n) for gcd(a, n) = 1, and Fermat's little theorem a^p−1 ≡ 1 (mod p) for a prime p not dividing a.

Converse fails: A₄ has order 12 but no subgroup of order 6.

Idea: Lagrange gives only a necessary condition on subgroup orders. Partial converses do hold: a cyclic group (more generally a finite abelian group) has a subgroup of every order dividing |G|, Cauchy's theorem gives subgroups of every prime order dividing |G|, and Sylow's theorem gives every prime-power order dividing |G|.

Counting with HK: for finite subgroups H, K of G, HK ⊆ G gives o(H)o(K)/o(H ∩ K) ≤ o(G). So if o(H)o(K) > o(G), then H ∩ K ≠ {e}.

Groups of order 4: up to isomorphism there are exactly two, the cyclic group ℤ₄ (some element has order 4) and the Klein four-group V₄ (every non-identity element has order 2).

Example 2: Show that every group of order 4 is abelian.

By Lagrange, each non-identity element has order 2 or 4. If some element has order 4, G is cyclic and hence abelian. Otherwise a² = e for all a, so a = a⁻¹ and ab = (ab)⁻¹ = b⁻¹a⁻¹ = ba.

Example 3: Find the remainder when 3¹⁰⁰ is divided by 7.

By Fermat, 3⁶ ≡ 1 (mod 7). Since 100 = 6·16 + 4, 3¹⁰⁰ ≡ 3⁴ = 81 ≡ 4 (mod 7).

9. Derivation: Cosets as Equivalence Classes

Step 1: for H ≤ G define a ~ b iff a⁻¹b ∈ H. It is reflexive (a⁻¹a = e ∈ H), symmetric (b⁻¹a = (a⁻¹b)⁻¹ ∈ H) and transitive (a⁻¹c = (a⁻¹b)(b⁻¹c) ∈ H).

Step 2: the class of a is {b : a⁻¹b ∈ H} = {ah : h ∈ H} = aH. Equivalence classes are equal or disjoint and cover G, which proves the 'equal or disjoint' and 'partition' properties at one stroke.

Step 3: h ↦ ah is a bijection H → aH (cancel a), so |aH| = |H|.

Step 4 (left and right cosets are equal in number): aH ↦ Ha⁻¹ is well defined and one-one, because aH = bH ⇔ a⁻¹b ∈ H ⇔ Ha⁻¹ = Hb⁻¹. It is onto, so the index [G : H] does not depend on the side, even though aH ≠ Ha in general.

10. Derivation: Counting Elements of Each Order in a Cyclic Group

Claim: in a cyclic group G of order n, for each divisor d of n there are exactly φ(d) elements of order d, and hence ∑_d|n φ(d) = n.

Proof: if o(x) = d, then ⟨x⟩ is a subgroup of order d, and G has only one such subgroup, H_d = ⟨a^n/d⟩. So the elements of order d are exactly the generators of the cyclic group H_d, and there are φ(d) of them (Standard Proofs, Theorem 16).

Every element has some order d dividing n, so summing over d counts G once: ∑_d|n φ(d) = n. Check for n = 12: φ(1) + φ(2) + φ(3) + φ(4) + φ(6) + φ(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12.

Use: this count answers 'how many elements of order 4 in ℤ₂₀?' at once: φ(4) = 2, namely 5 and 15.

11. Choosing the Method

If you must show H ≤ G for an infinite G → check H ≠ ∅ (usually e ∈ H) and use the one-step test ab⁻¹ ∈ H.

If H is finite → closure alone suffices (finite subgroup test).

If asked whether G is cyclic → for 'yes' exhibit an element of order |G|; for 'no' show every element has order less than |G|, or that G is non-abelian, or that G has two different subgroups of the same order.

If the question is about possible orders of elements or subgroups of a finite group → Lagrange first, then o(a^k) = n/gcd(n, k) and the φ(d) count.

If a remainder of a large power is asked → Fermat or Euler, i.e. Lagrange in U(n).

If you must prove G abelian → try a² = e for all a, (ab)² = a²b², o(G) = p or p², or G/Z(G) cyclic.

If you must show H ∩ K ≠ {e} → count: |H||K|/|H ∩ K| = |HK| ≤ |G|.

12. Common Mistakes

Using the finite subgroup test on an infinite set: ℕ ⊆ ℤ is closed under + but is not a subgroup. Use the one-step test unless H is known to be finite.

Assuming the converse of Lagrange: '6 divides 12, so A₄ has a subgroup of order 6' is false. Existence of subgroups needs Cauchy, Sylow or cyclicity.

Writing (ab)⁻¹ = a⁻¹b⁻¹: the correct rule is (ab)⁻¹ = b⁻¹a⁻¹; the two agree only when a and b commute.

Assuming o(ab) = lcm(o(a), o(b)): this needs ab = ba (and o(ab) = o(a)o(b) needs, in addition, gcd(o(a), o(b)) = 1). In S₃, (1 2) and (1 3) have order 2 but their product (1 3 2) has order 3.

Treating HK as a subgroup: HK is a subgroup only when HK = KH (for example when one of them is normal); the product formula counts HK as a set either way.

Listing too many generators of ℤ: an infinite cyclic group has exactly two generators, ±1 in ℤ; the φ(n) rule is only for finite cyclic groups.

13. Connections

Link: cosets are the group version of the solution set of a linear system: the solutions of Ax = b form x₀ + N(A), a coset of the null space in Linear Algebra.

Link: the vector space axioms of Linear Algebra begin with '(V, +) is an abelian group', and the multiplicative group of a finite field (Field Extensions topic) is cyclic, which is why primitive roots mod p exist.

14. Worked Examples

Example 4: Show that every group of even order contains an element of order 2.

Pair each element a with its inverse a⁻¹. The elements with a ≠ a⁻¹ fall into disjoint pairs {a, a⁻¹}, so their number is even.

Hence the number of elements with a = a⁻¹, i.e. a² = e, is o(G) minus an even number, which is even.

This set contains e, so it contains at least one more element a ≠ e, and that element has order 2.

Example 5: Prove or disprove: U(8) is cyclic.

U(8) = {1, 3, 5, 7} under multiplication mod 8 is a group of order 4.

3² = 9 ≡ 1, 5² = 25 ≡ 1 and 7² = 49 ≡ 1 (mod 8), so every non-identity element has order 2.

A cyclic group of order 4 needs an element of order 4, so U(8) is not cyclic; it is isomorphic to the Klein four-group V₄.

Example 6: Let G be a group such that (ab)ⁿ = aⁿbⁿ for all a, b ∈ G and for three consecutive integers n = k, k + 1, k + 2. Show that G is abelian.

Writing (ab)^k+1 = (ab)^k(ab) gives a^k+1b^k+1 = a^kb^kab. Cancel a^k on the left and b on the right: ab^k = b^ka.

In the same way, n = k + 1 and n = k + 2 give ab^k+1 = b^k+1a.

Now ab^k+1 = (ab^k)b = b^kab, while b^k+1a = b^k(ba). Cancelling b^k on the left gives ab = ba, so G is abelian.

15. Solved Past UPSC Questions

Example 7 (UPSC 2015, 10 marks): (i) How many generators are there of the cyclic group G of order 8? Explain.
(ii) Taking a group {e, a, b, c} of order 4, where e is the identity, construct composition tables showing that one is cyclic while the other is not. (5 + 5 = 10)

(i) If G = ⟨a⟩ has order 8, then o(a^k) = 8/gcd(8, k), so a^k generates G iff gcd(k, 8) = 1. The generators are a, a³, a⁵, a⁷: there are φ(8) = 4 of them.

(ii) Cyclic table: take o(a) = 4, b = a², c = a³. Rows (row element × column element, columns e, a, b, c): e: e, a, b, c; a: a, b, c, e; b: b, c, e, a; c: c, e, a, b.

This group is generated by a (and by c), so it is cyclic and isomorphic to ℤ₄.

Non-cyclic table: take a² = b² = c² = e, ab = ba = c, bc = cb = a, ca = ac = b. Rows: e: e, a, b, c; a: a, e, c, b; b: b, c, e, a; c: c, b, a, e.

This is the group ℤ₂ × ℤ₂ (a = (1, 0), b = (0, 1), c = (1, 1)), so associativity holds; each row and column is a permutation of {e, a, b, c}. Every non-identity element has order 2, so no element generates the group: it is not cyclic.

Answer: (i) 4 generators, a, a³, a⁵, a⁷; (ii) the first table is the cyclic group ℤ₄, the second the non-cyclic Klein group V₄.

Example 8 (UPSC 2018, 20 marks): Find all the proper subgroups of the multiplicative group of the field (ℤ₁₃, +₁₃, ×₁₃), where +₁₃ and ×₁₃ represent addition modulo 13 and multiplication modulo 13 respectively.

The multiplicative group is U(13) = {1, 2, ..., 12}, of order 12. Powers of 2 mod 13 are 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1, so o(2) = 12 and U(13) = ⟨2⟩ is cyclic.

A cyclic group of order 12 has exactly one subgroup of each order d dividing 12, namely ⟨2^12/d⟩.

d = 1: {1}; d = 2: ⟨2⁶⟩ = ⟨12⟩ = {1, 12}; d = 3: ⟨2⁴⟩ = ⟨3⟩ = {1, 3, 9}.

d = 4: ⟨2³⟩ = ⟨8⟩ = {1, 8, 12, 5}; d = 6: ⟨2²⟩ = ⟨4⟩ = {1, 4, 3, 12, 9, 10}; d = 12 gives U(13) itself.

Answer: the proper subgroups are {1}, {1, 12}, {1, 3, 9}, {1, 5, 8, 12} and {1, 3, 4, 9, 10, 12} (the last four are the non-trivial ones).

Example 9 (UPSC 2019, 10 marks): Let G be a finite group, H and K subgroups of G such that K ⊂ H. Show that (G : K) = (G : H)(H : K).

G is finite, so by Lagrange's theorem (G : K) = o(G)/o(K), (G : H) = o(G)/o(H) and (H : K) = o(H)/o(K).

Hence (G : H)(H : K) = (o(G)/o(H))·(o(H)/o(K)) = o(G)/o(K) = (G : K).

(A coset proof works for finite indices in any group: if G is the disjoint union of a_iH (1 ≤ i ≤ m) and H the disjoint union of b_jK (1 ≤ j ≤ n), then G is the disjoint union of the mn cosets a_ib_jK.)

Answer: (G : K) = (G : H)(H : K).

Example 10 (UPSC 2021, 15 marks): Show that there are infinitely many subgroups of the additive group ℚ of rational numbers.

For each n ∈ ℕ, H_n = nℤ = {nk : k ∈ ℤ} is a subgroup of (ℚ, +): it is non-empty and nk − nl = n(k − l) ∈ H_n.

If m < n, then m ∈ H_m but m ∉ H_n, because every positive element of H_n is at least n. So the subgroups H₁, H₂, H₃, ... are pairwise distinct.

Similarly ⟨1/n⟩ = {k/n : k ∈ ℤ}, n ≥ 1, are infinitely many distinct subgroups not contained in ℤ. Note that ℚ itself is not cyclic: if ℚ = ⟨r⟩ with r ≠ 0, then r/2 would be an integer multiple of r, which is false.

Answer: the subgroups nℤ (n = 1, 2, 3, ...) are infinitely many distinct subgroups of (ℚ, +).

Example 11 (UPSC 2025, 10 marks): Let H and K be two subgroups of a group G such that o(H) > √o(G) and o(K) > √o(G). Show that H ∩ K ≠ {e}, where e is the identity element. Here o(H), o(K) and o(G) denote the order of H, K and G respectively.

HK = {hk : h ∈ H, k ∈ K} is a subset of G (not necessarily a subgroup), and |HK| = o(H)o(K)/o(H ∩ K) for finite subgroups H, K.

Suppose H ∩ K = {e}. Then |HK| = o(H)o(K) > √o(G)·√o(G) = o(G).

But HK ⊆ G, so |HK| ≤ o(G), a contradiction.

Answer: H ∩ K ≠ {e}; H and K have a common non-identity element.

16. Practice Exercises

Exercise 1 (basic): On ℤ define a ∗ b = a + b − 1. Show that (ℤ, ∗) is a group and find the identity and the inverse of a.

Answer: closure is clear; (a ∗ b) ∗ c = a + b + c − 2 = a ∗ (b ∗ c); the identity is 1 (a ∗ 1 = a); the inverse of a is 2 − a. The group is abelian.

Exercise 2 (basic): Find the order of every element of U(10) = {1, 3, 7, 9} and decide whether U(10) is cyclic.

Answer: o(1) = 1, o(9) = 2, o(3) = o(7) = 4 (3, 9, 7, 1 are the powers of 3). So U(10) = ⟨3⟩ = ⟨7⟩ is cyclic.

Exercise 3 (basic): List the cosets of H = {0, 5, 10, 15} in ℤ₂₀ and state the index [ℤ₂₀ : H].

Answer: {0, 5, 10, 15}, {1, 6, 11, 16}, {2, 7, 12, 17}, {3, 8, 13, 18}, {4, 9, 14, 19}; the index is 20/4 = 5.

Exercise 4 (standard): How many elements of order 6 does ℤ₃₆ have? List them.

Answer: φ(6) = 2 elements, namely 6 and 30 (36/gcd(36, k) = 6 forces gcd(36, k) = 6).

Exercise 5 (standard): Find all subgroups of ℤ₁₈, with their orders, and say which contain which.

Answer: one for each divisor: ⟨0⟩ (1), ⟨9⟩ = {0, 9} (2), ⟨6⟩ = {0, 6, 12} (3), ⟨3⟩ (6), ⟨2⟩ (9), ℤ₁₈ (18). The subgroup of order d lies in that of order e iff d | e.

Exercise 6 (standard): Find the remainder when 7²²² is divided by 15.

Answer: 4. Here φ(15) = 8 and 222 = 8·27 + 6; in fact 7⁴ ≡ 1, so 7²²² ≡ 7² = 49 ≡ 4 (mod 15).

Exercise 7 (standard): Let G be an abelian group. Show that H = {a ∈ G : a² = e} is a subgroup, and show by an instance in S₃ that this can fail when G is not abelian.

Answer: e ∈ H, and for a, b ∈ H, (ab⁻¹)² = a²(b²)⁻¹ = e because G is abelian. In S₃, (1 2) and (1 3) square to e but (1 2)(1 3) = (1 3 2) has order 3.

Exercise 8 (UPSC level): H and K are subgroups of a group G of order 72 with o(H) = 12 and o(K) = 18. Find the possible values of o(H ∩ K).

Answer: 3 or 6. o(H ∩ K) divides gcd(12, 18) = 6 by Lagrange, and |HK| = 216/o(H ∩ K) ≤ 72 forces o(H ∩ K) ≥ 3.

Exercise 9 (UPSC level): Let G be a finite abelian group of order n and m an integer with gcd(m, n) = 1. Prove that x ↦ x^m is a one-one map of G onto G, so every element of G has a unique m-th root.

Answer: since G is abelian, x^my^−m = (xy⁻¹)^m; if x^m = y^m, then o(xy⁻¹) divides both m and n (Lagrange), so it is 1 and x = y. A one-one map of a finite set to itself is onto.

Exercise 10 (UPSC level): Prove that (ℚ, +) is not cyclic, but every finitely generated subgroup of (ℚ, +) is cyclic.

Answer: if ℚ = ⟨r⟩ with r ≠ 0, then r/2 = kr for an integer k, which is impossible. For ⟨a₁/b, ..., a_k/b⟩ (common denominator b) every generator lies in the cyclic group ⟨1/b⟩, and a subgroup of a cyclic group is cyclic.

1. Understanding Normal Subgroups

Why we need it: we want to 'divide' a group G by a subgroup H and make the cosets of H into a new group. For that, the product of two cosets must not depend on which representatives we pick. This works only for special subgroups, the normal ones.

In words: gHg⁻¹ = {ghg⁻¹ : h ∈ H} is H 'seen through g': take each h and conjugate it by g. H is normal (H ⊴ G) if gHg⁻¹ ⊆ H for every g ∈ G.

'For every g' is the demanding part: a single g with one ghg⁻¹ outside H destroys normality. The equivalent form gH = Hg is an equality of sets, not gh = hg for each h.

First example, slowly: let A₃ = {e, (1 2 3), (1 3 2)} in S₃ and take g = (1 2). Conjugating a cycle by g renames its symbols by g (the relabelling rule of the Permutation Groups topic).

Step 1: (1 2)(1 2 3)(1 2)⁻¹ = (2 1 3) = (1 3 2) ∈ A₃. Step 2: (1 2)(1 3 2)(1 2)⁻¹ = (2 3 1) = (1 2 3) ∈ A₃. Step 3: geg⁻¹ = e.

Step 4: the same happens for g = (1 3) and g = (2 3), and conjugation by an element of the abelian group A₃ fixes each element of A₃. So gA₃g⁻¹ ⊆ A₃ for all six g, and A₃ ⊴ S₃.

What to notice: a quicker route is that [S₃ : A₃] = 2, and every subgroup of index 2 is normal.

Non-example: H = {e, (1 2)} in S₃ with g = (1 3): g(1 2)g⁻¹ = (3 2) ∉ H, so H is not normal. Correspondingly (1 3)H = {(1 3), (1 2 3)} differs from H(1 3) = {(1 3), (1 3 2)}.

Building up: the standard tests, in the order to try them: every subgroup of an abelian group is normal (ghg⁻¹ = h); a subgroup of index 2 is normal (Standard Proofs, Theorem 4); a kernel is normal (Theorem 5).

Also: the only subgroup of a given order is normal (Example 3 below), and the centre Z(G) is normal. Only if none of these applies, check gng⁻¹ ∈ N directly; the 'Why ⊆ suffices' line shows that one inclusion is enough.

Where it is used: quotient groups, simple groups, Sylow theory (n_p = 1 exactly when the Sylow p-subgroup is normal) and internal direct products.

2. Understanding Quotient Groups

Why we need it: arithmetic mod n is the integers with multiples of n ignored. A quotient group does the same for any group: it collapses a normal subgroup N to the identity and keeps the rest of the structure.

In words: the elements of G/N are cosets gN, which are sets. To multiply two cosets, pick a representative of each, multiply them in G and take the coset of the product: (aN)(bN) = abN.

The identity of G/N is eN = N itself, the inverse of aN is a⁻¹N, and for finite G the order is |G/N| = |G|/|N|.

First example, slowly: take G = ℤ₁₂ and N = ⟨4⟩ = {0, 4, 8}. N is normal because G is abelian.

Step 1 (the elements): N, 1 + N = {1, 5, 9}, 2 + N = {2, 6, 10}, 3 + N = {3, 7, 11}. So G/N has 12/3 = 4 elements.

Step 2 (a sum): (2 + N) + (3 + N) = 5 + N, and 5 ∈ 1 + N, so the sum is 1 + N.

Step 3 (independence of representatives): use 6 ∈ 2 + N and 11 ∈ 3 + N instead. Then 6 + 11 = 17 ≡ 5, again in 1 + N. The answer did not change.

Step 4 (identify the group): the multiples of 1 + N are 1 + N, 2 + N, 3 + N, N, so 1 + N has order 4 and G/N ≅ ℤ₄, in line with ℤ_n/⟨d⟩ ≅ ℤ_d.

Non-example: for H = {e, (1 2)} in S₃, the 'product' of cosets depends on the representatives (see the derivation 'Coset Multiplication is Well Defined iff N is Normal' below), so S₃/H is not a group.

Building up: (1) Normality of N makes the product well defined (that derivation). (2) The axioms are then inherited from G: (aN·bN)·cN = (ab)cN = a(bc)N = aN·(bN·cN).

(3) |G/N| = [G : N] = |G|/|N| by Lagrange. (4) Properties pass down: if G is abelian so is G/N, and if G = ⟨g⟩ then G/N = ⟨gN⟩.

Where it is used: the first isomorphism theorem, the G/Z(G) argument for abelian groups, the construction ℤ_n = ℤ/nℤ, and quotient rings R/I in the Rings topic.

3. Understanding Homomorphisms and Kernels

Why we need it: to compare two groups we need maps that respect the operations. Such a map carries information from one group to the other; when it is also a bijection the two groups are 'the same group with different names'.

In words: f : G → G′ is a homomorphism if f(ab) = f(a)f(b) for all a, b. On the left the product ab is computed in G; on the right f(a)f(b) is computed in G′.

The kernel ker f is the set of elements sent to the identity e′ of G′; it measures how much f collapses. The image f(G) is the set of values actually taken.

First example, slowly: let f : ℤ → ℤ₅, f(n) = n mod 5.

Step 1: f(m + n) = (m + n) mod 5 = (m mod 5 + n mod 5) mod 5 = f(m) + f(n), so f is a homomorphism of additive groups.

Step 2: ker f = {n : 5 | n} = 5ℤ. Step 3: f(k) = k for k = 0, 1, 2, 3, 4, so f is onto.

Step 4: f(a) = f(b) exactly when a − b ∈ 5ℤ, i.e. when a and b lie in the same coset of the kernel. So f is not one-one: f(2) = f(7).

What to notice: the points with the same image form a coset of the kernel. This is the whole idea behind the first isomorphism theorem.

Non-example: g : ℤ → ℤ, g(n) = n + 1, is not a homomorphism: g(0 + 0) = 1 but g(0) + g(0) = 2. A quick test is that a homomorphism must send the identity to the identity, and g(0) = 1 ≠ 0.

Building up: (1) f(e) = e′: f(e) = f(ee) = f(e)f(e), then cancel. (2) f(a⁻¹) = f(a)⁻¹, since f(a)f(a⁻¹) = f(e) = e′.

(3) If aⁿ = e then f(a)ⁿ = f(aⁿ) = e′, so o(f(a)) divides o(a). (4) ker f is a normal subgroup, and f is one-one iff ker f = {e} (Standard Proofs, Theorem 5).

Where it is used: proving normality by exhibiting a kernel, counting homomorphisms ℤ_m → ℤ_n through f(1), proving two groups are not isomorphic, and every quotient identification.

4. Understanding the First Isomorphism Theorem

Why we need it: a quotient G/N is a group of cosets, which is hard to picture. The theorem lets us recognise it as a familiar group: find an onto homomorphism whose kernel is exactly N.

In words: G/ker f ≅ f(G), and the isomorphism sends the coset a·ker f to f(a). Collapse exactly what f collapses, and what remains is a copy of the image.

First example, slowly: let ℂ^* be the non-zero complex numbers under multiplication and T = {z : |z| = 1} the unit circle. We show ℂ^*/T ≅ ℝ⁺.

Step 1 (the map): f(z) = |z|, from ℂ^* to ℝ⁺ (positive reals under multiplication). Step 2 (homomorphism): |zw| = |z||w|.

Step 3 (onto): for r > 0, f(r) = r. Step 4 (kernel): f(z) = 1 iff |z| = 1, so ker f = T.

Step 5 (conclusion): by the theorem, ℂ^*/T ≅ ℝ⁺. Picture: the coset zT is the circle of radius |z| about 0, and multiplying two such circles multiplies their radii.

Non-example: to identify ℤ/4ℤ one might try n ↦ n mod 2 onto ℤ₂. Its kernel is 2ℤ, not 4ℤ, so the theorem gives ℤ/2ℤ ≅ ℤ₂ and says nothing about ℤ/4ℤ. The kernel must equal N exactly.

Building up: put K = ker f and define φ(aK) = f(a). (1) aK = bK ⇔ a⁻¹b ∈ K ⇔ f(a) = f(b): read '⇒' as 'φ is well defined' and '⇐' as 'φ is one-one'.

(2) φ(aK·bK) = f(ab) = f(a)f(b), so φ is a homomorphism. (3) φ is onto f(G) by definition. Full proof: Standard Proofs, Theorem 6; in practice use the five-step method below.

Where it is used: every 'show G/N ≅ G′' question, the counting of homomorphisms, the ring version R/ker f ≅ f(R), and rank–nullity in Linear Algebra.

5. Normal Subgroups

Definition: H ≤ G is normal, written H ⊴ G, if gHg⁻¹ ⊆ H for all g ∈ G; equivalently gH = Hg for all g ∈ G.

Idea: conjugation x ↦ gxg⁻¹ is a relabelling of G by g, and a normal subgroup is one that no relabelling can move. Note that gH = Hg is an equality of sets, not gh = hg for each h: in S₃, (1 2)(1 2 3) = (2 3) but (1 2 3)(1 2) = (1 3), yet (1 2)A₃ = A₃(1 2), both being the set of odd permutations.

Why ⊆ suffices: if gHg⁻¹ ⊆ H for every g, apply it to g⁻¹ to get g⁻¹Hg ⊆ H, i.e. H ⊆ gHg⁻¹. So the inclusion for all g already gives gHg⁻¹ = H.

Quick tests: every subgroup of an abelian group is normal; every subgroup of index 2 is normal; the centre Z(G) is normal.

Results: an intersection of normal subgroups is normal; if N ⊴ G and H ≤ G then HN is a subgroup; if H, N are both normal then HN is normal.

Not transitive: in D₄ = ⟨r, s⟩, {e, s} ⊴ {e, s, r², sr²} ⊴ D₄, but {e, s} is not normal in D₄.

Idea: normality is relative to the big group. K ⊴ H only controls conjugation by elements of H; an element of G outside H can move K, as the rotation r conjugates {e, s} to {e, sr²} in D₄.

Simple group: a group whose only normal subgroups are {e} and G; ℤ_p (p prime) and A_n (n ≥ 5) are simple.

Idea: simple groups are the 'primes' of group theory: a group with a proper non-trivial normal subgroup N can be broken into the smaller pieces N and G/N, and a simple group cannot.

6. Quotient Groups

Definition: if N ⊴ G, the set G/N = {gN : g ∈ G} is a group under (aN)(bN) = abN, with identity N and inverse (aN)⁻¹ = a⁻¹N.

Idea: G/N is G seen with blurred vision: the elements of N become invisible (they all act as the identity), and two elements are identified when they differ by an element of N. ℤ/nℤ is 'the integers when only the remainder mod n is visible'.

Why normality is needed: the rule aN·bN = abN is well-defined (independent of the representatives a, b) exactly when N is normal.

Order: if G is finite, o(G/N) = o(G)/o(N).

Results: a quotient of an abelian group is abelian and a quotient of a cyclic group is cyclic; if G/Z(G) is cyclic then G is abelian.

Quotients of cyclic groups: the subgroups of ℤ_n are ⟨d⟩ for d | n, with o(⟨d⟩) = n/d, and ℤ_n/⟨d⟩ ≅ ℤ_d. Also ℤ/nℤ ≅ ℤ_n, and ℝ/ℤ ≅ T = {z ∈ ℂ : |z| = 1} via t ↦ e^2πit.

Commutator subgroup: G′ is generated by all aba⁻¹b⁻¹; G′ ⊴ G, and G/N is abelian iff G′ ⊆ N.

7. Homomorphisms

Homomorphism: a map f : G → G′ with f(ab) = f(a)f(b) for all a, b ∈ G.

Idea: a homomorphism respects the multiplication table: it may collapse many elements to one (the kernel measures how much), but it never scrambles products. Standard examples: det : GL(n, ℝ) → ℝ^*, sgn : S_n → {±1}, exp : (ℝ, +) → (ℝ⁺, ·), and reduction mod n : ℤ → ℤ_n.

Properties: f(e) = e′; f(a⁻¹) = f(a)⁻¹; if o(a) is finite then o(f(a)) divides o(a); images and inverse images of subgroups are subgroups.

Order restrictions for finite groups: if f : G → G′ is a homomorphism of finite groups, o(f(G)) divides o(G) (since f(G) ≅ G/ker f) and o(G′) (Lagrange), hence divides gcd(o(G), o(G′)). If this gcd is 1, f is trivial; an onto homomorphism needs o(G′) | o(G).

Images inherit properties: a homomorphic image of an abelian group is abelian, and of a cyclic group is cyclic (f(⟨a⟩) = ⟨f(a)⟩); the converses fail.

Normality via kernels: to show N ⊴ G it is often quickest to exhibit a homomorphism from G whose kernel is N (for example det, sign, or 'read off the diagonal entries').

Kernel: ker f = {a ∈ G : f(a) = e′} is a normal subgroup of G, and f is one-one iff ker f = {e}.

Why it works: f(a) = f(b) ⇔ f(a⁻¹b) = e′ ⇔ a⁻¹b ∈ ker f ⇔ a ker f = b ker f. So the set of points with the same image as a is exactly the coset a·ker f, and f is one-one iff these cosets are single points, i.e. ker f = {e}.

Isomorphism: a bijective homomorphism, written G ≅ G′. An automorphism is an isomorphism of G onto itself; the inner automorphisms are x ↦ gxg⁻¹, and Inn(G) ≅ G/Z(G).

Natural map: π : G → G/N, π(g) = gN, is an onto homomorphism with kernel N, so every normal subgroup is the kernel of some homomorphism.

8. Isomorphism Theorems

First (fundamental) theorem: if f : G → G′ is a homomorphism, then G/ker f ≅ f(G). In particular, if f is onto then G/ker f ≅ G′.

Idea: every homomorphic image is a quotient. f squashes each coset of ker f to one point, and these points are exactly f(G); so up to isomorphism the images of G are the groups G/N, N ⊴ G. The proof is in Standard Proofs, Theorem 6.

Second theorem: if H ≤ G and N ⊴ G, then H ∩ N ⊴ H and HN/N ≅ H/(H ∩ N).

Third theorem: if N ⊴ G, K ⊴ G and N ⊆ K, then K/N ⊴ G/N and (G/N)/(K/N) ≅ G/K.

Correspondence theorem: subgroups of G/N correspond one-to-one with subgroups of G containing N, and normal subgroups correspond to normal subgroups.

Idea: the subgroup lattice of G/N is the part of the lattice of G lying above N. For example the subgroups of ℤ/12ℤ correspond to the subgroups dℤ of ℤ that contain 12ℤ, i.e. to the divisors d of 12.

Method for proving G/N ≅ G′:

1. Define a map f from G to G′ suggested by the problem.

2. Prove that f(ab) = f(a)f(b).

3. Prove that f is onto.

4. Compute ker f and show that it equals N.

5. Conclude G/N ≅ G′ by the first isomorphism theorem.

Example 1: Show that ℝ^*/{1, −1} ≅ ℝ⁺, both under multiplication.

Define f(x) = |x|. Then f(xy) = |xy| = |x||y| = f(x)f(y), f is onto ℝ⁺ since f(r) = r for r > 0, and ker f = {x : |x| = 1} = {1, −1}. Hence ℝ^*/{1, −1} ≅ ℝ⁺.

Example 2: Show that SL(n, ℝ) ⊴ GL(n, ℝ) and GL(n, ℝ)/SL(n, ℝ) ≅ ℝ^*.

Take f(A) = det A. Then det(AB) = det A·det B, f is onto because diag(a, 1, ..., 1) has determinant a, and ker f = {A : det A = 1} = SL(n, ℝ). So SL(n, ℝ) is normal and the quotient is ≅ ℝ^*.

9. Derivation: Coset Multiplication is Well Defined iff N is Normal

(⇐) Let N ⊴ G, aN = a′N and bN = b′N. Then a′ = an₁ and b′ = bn₂ with n₁, n₂ ∈ N, and a′b′ = an₁bn₂ = ab(b⁻¹n₁b)n₂. Since b⁻¹n₁b ∈ N, a′b′ ∈ abN, so a′b′N = abN.

(⇒) Suppose the rule is well defined, and let n ∈ N, g ∈ G. Since nN = eN, the products (nN)(g⁻¹N) and (eN)(g⁻¹N) agree: ng⁻¹N = g⁻¹N. Hence (g⁻¹)⁻¹ng⁻¹ = gng⁻¹ ∈ N, so N is normal.

Moral: normal subgroups are exactly the subgroups that can be 'divided out'. With H = {e, (1 2)} in S₃, H = (1 2)H, yet multiplying each on the right by (1 3)H would give (1 3)H = {(1 3), (1 2 3)} and (1 2)(1 3)H = (1 3 2)H = {(1 3 2), (2 3)}, which are different cosets.

10. Derivation: Second and Third Isomorphism Theorems

Second theorem: let H ≤ G and N ⊴ G. Then HN is a subgroup (HN = NH because N is normal) and N ⊴ HN. Define φ : H → HN/N by φ(h) = hN; it is the natural map restricted to H, so it is a homomorphism.

φ is onto, because a coset hnN of HN/N equals hN = φ(h). Its kernel is {h ∈ H : hN = N} = H ∩ N. By the first theorem H ∩ N ⊴ H and H/(H ∩ N) ≅ HN/N; for finite groups this re-derives |HN| = |H||N|/|H ∩ N|.

Third theorem: let N ⊆ K with N, K ⊴ G. Define ψ : G/N → G/K by ψ(gN) = gK. It is well defined, since gN = g′N ⇒ g⁻¹g′ ∈ N ⊆ K ⇒ gK = g′K; it is clearly an onto homomorphism.

ker ψ = {gN : gK = K} = {gN : g ∈ K} = K/N. So K/N ⊴ G/N and (G/N)/(K/N) ≅ G/K: 'the N cancels'. Example: (ℤ/12ℤ)/(4ℤ/12ℤ) ≅ ℤ/4ℤ.

11. Derivation: Number of Homomorphisms ℤ_m → ℤ_n

A homomorphism f is fixed by a = f(1), since f(k) = ka. The rule k ↦ ka (mod n) is well defined on ℤ_m iff ma ≡ 0 (mod n), i.e. iff o(a) divides m.

With d = gcd(m, n): n | ma ⇔ (n/d) | (m/d)a ⇔ (n/d) | a. The multiples of n/d in ℤ_n are d in number, so there are exactly gcd(m, n) homomorphisms, f(k) = j(n/d)k for j = 0, 1, ..., d − 1.

Check: ℤ₁₂ → ℤ₃₀ has gcd(12, 30) = 6 homomorphisms (Example 5 below), and ℤ_m → ℤ_n has only the trivial one when gcd(m, n) = 1.

12. Choosing the Method

If you must show N ⊴ G → in this order try: G abelian; [G : N] = 2; N is the kernel of a natural homomorphism (det, sgn, a projection, 'read off the diagonal'); N is the only subgroup of its order; finally the direct check gng⁻¹ ∈ N.

If you must show N is not normal → one explicit pair g ∈ G, n ∈ N with gng⁻¹ ∉ N is enough.

If you must identify G/N → find an onto homomorphism with kernel N and use the first theorem (the five-step method above); for a small finite quotient, its order and element orders often decide it.

If you must count or find homomorphisms G → G′ → send generators to elements whose orders divide theirs (and divide |G′|), and check the defining relations.

If asked whether an onto homomorphism G → G′ exists → |G′| must divide |G|, and G must have a normal subgroup N with G/N ≅ G′.

If you must decide G ≅ G′ → for 'no', compare invariants: order, abelian or not, cyclic or not, number of elements of each order, the centre, solvability of equations such as x² = a (the method of Example 6).

13. Common Mistakes

Checking gn = ng instead of gNg⁻¹ = N: elementwise commuting is much stronger and usually false; only the set equality gN = Ng is required.

Forming G/H for a non-normal H: S₃/{e, (1 2)} is not a group (see the derivation above). Always state 'N is normal, so G/N is a group'.

Defining a map on cosets without checking it is well defined: k ↦ k mod 5 from ℤ₁₂ to ℤ₅ is not a function, since 0 = 12 in ℤ₁₂ but 0 ≠ 2 in ℤ₅.

Skipping 'onto': without it the first theorem gives only G/ker f ≅ f(G), not ≅ G′.

Lifting properties from a quotient: G/N abelian (or cyclic) does not make G abelian; S₃/A₃ ≅ ℤ₂. The one exception to remember is N = Z(G) with G/Z(G) cyclic.

14. Connections

Link: the rank–nullity theorem of Linear Algebra is the first isomorphism theorem for vector spaces: V/ker T ≅ T(V), so dim V − nullity T = rank T.

Link: the same three theorems hold for rings, with ideals in place of normal subgroups (Rings topic); and ℝ/ℤ ≅ unit circle via t ↦ e^2πit is the algebra behind periodic functions and Fourier series (PDE chapter).

15. Worked Examples

Example 3: If H is the only subgroup of order m in a group G, show that H ⊴ G.

For g ∈ G, the map x ↦ gxg⁻¹ is an automorphism of G, so gHg⁻¹ is a subgroup of G.

This map is one-one, so o(gHg⁻¹) = o(H) = m. By uniqueness gHg⁻¹ = H for every g ∈ G, hence H ⊴ G.

Application: A₃ = ⟨(1 2 3)⟩ is the only subgroup of order 3 in S₃, so it is normal in S₃.

Example 4: Prove that if G/Z(G) is cyclic, then G is abelian.

Write Z = Z(G) and let G/Z = ⟨gZ⟩. Every coset is g^mZ for some integer m, so every a ∈ G has the form a = g^mz with z ∈ Z.

Take a = g^mz₁ and b = gⁿz₂. Since z₁, z₂ commute with every element, ab = g^m+nz₁z₂ = g^n+mz₂z₁ = ba.

Hence G is abelian; then Z(G) = G, so in fact G/Z(G) is trivial. Thus G/Z(G) can never be a non-trivial cyclic group.

Example 5: Find all group homomorphisms from ℤ₁₂ to ℤ₃₀.

A homomorphism f is fixed by a = f(1), since f(x) = xa (mod 30). This rule is well defined iff 12a ≡ 0 (mod 30).

30 | 12a ⇔ 5 | 2a ⇔ 5 | a, so a ∈ {0, 5, 10, 15, 20, 25}.

Hence there are exactly gcd(12, 30) = 6 homomorphisms, f(x) = 5kx (mod 30) for k = 0, 1, 2, 3, 4, 5; none is onto, since the image has at most 12 elements and ℤ₃₀ has 30.

Example 6: Prove or disprove: (ℚ, +) ≅ (ℚ⁺, ·).

Suppose f : ℚ → ℚ⁺ is an isomorphism. Since f is onto, f(a) = 2 for some a ∈ ℚ.

Then 2 = f(a) = f(a/2 + a/2) = f(a/2)², so r = f(a/2) is a positive rational number with r² = 2.

This is impossible because √2 is irrational. Hence (ℚ, +) and (ℚ⁺, ·) are not isomorphic.

16. Solved Past UPSC Questions

Example 7 (UPSC 2014, 10 marks): Let G be the set of all real 2 × 2 matrices [[x, y], [0, z]], where xz ≠ 0. Show that G is a group under matrix multiplication. Let N denote the subset {[[1, a], [0, 1]] : a ∈ ℝ}. Is N a normal subgroup of G? Justify your answer.

Closure: [[x, y], [0, z]]·[[x′, y′], [0, z′]] = [[xx′, xy′ + yz′], [0, zz′]], and (xx′)(zz′) = (xz)(x′z′) ≠ 0.

Matrix multiplication is associative; I = [[1, 0], [0, 1]] ∈ G; and [[1/x, −y/(xz)], [0, 1/z]] ∈ G is the inverse of [[x, y], [0, z]]. So G is a (non-abelian) group.

Define φ : G → ℝ^* × ℝ^* by φ([[x, y], [0, z]]) = (x, z). By the product formula φ(AB) = (xx′, zz′) = φ(A)φ(B), so φ is a homomorphism.

ker φ = {A ∈ G : x = z = 1} = {[[1, a], [0, 1]] : a ∈ ℝ} = N. The kernel of a homomorphism is a normal subgroup, so N ⊴ G.

Direct check: for A = [[x, y], [0, z]], A[[1, a], [0, 1]]A⁻¹ = [[1, xa/z], [0, 1]] ∈ N.

Answer: G is a group and N is a normal subgroup of G (the kernel of A ↦ (x, z)); moreover G/N ≅ ℝ^* × ℝ^*, since φ is onto.

Example 8 (UPSC 2018, 15 marks): Show that the quotient group of (ℝ, +) modulo ℤ is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here ℝ is the set of real numbers and ℤ is the set of integers.

Let T = {z ∈ ℂ : |z| = 1} under multiplication, and define φ : ℝ → T by φ(t) = e^2πit = cos 2πt + i sin 2πt.

φ(s + t) = e^2πi(s+t) = e^2πise^2πit = φ(s)φ(t), so φ is a homomorphism from (ℝ, +) to (T, ·).

Onto: every z ∈ T equals e^iθ for some θ ∈ [0, 2π), and then z = φ(θ/(2π)).

Kernel: e^2πit = 1 ⇔ cos 2πt = 1 and sin 2πt = 0 ⇔ t ∈ ℤ, so ker φ = ℤ.

By the first isomorphism theorem ℝ/ℤ ≅ T.

Answer: ℝ/ℤ ≅ T, the isomorphism being t + ℤ ↦ e^2πit.

Example 9 (UPSC 2019, 10 marks): If G and H are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from G to H, the trivial one.

Let f : G → H be a homomorphism with kernel K. By the first isomorphism theorem G/K ≅ f(G), so o(f(G)) = o(G)/o(K) divides o(G).

f(G) is a subgroup of H, so by Lagrange o(f(G)) divides o(H).

Hence o(f(G)) divides gcd(o(G), o(H)) = 1, so f(G) = {e_H}: f(g) = e_H for every g ∈ G.

The trivial map g ↦ e_H is always a homomorphism, so it is the only one.

Answer: the only homomorphism G → H is the trivial one.

Example 10 (UPSC 2019, 10 marks): Write down all quotient groups of the group Z₁₂.

ℤ₁₂ is cyclic, hence abelian, so every subgroup is normal. Its subgroups are ⟨d⟩ for d = 1, 2, 3, 4, 6, 12 (⟨12⟩ = ⟨0⟩), of orders 12, 6, 4, 3, 2, 1.

ℤ₁₂/⟨0⟩ ≅ ℤ₁₂, and ℤ₁₂/⟨1⟩ = {ℤ₁₂} is the trivial group.

ℤ₁₂/⟨2⟩ = {⟨2⟩, 1 + ⟨2⟩} ≅ ℤ₂, where ⟨2⟩ = {0, 2, 4, 6, 8, 10}.

ℤ₁₂/⟨3⟩ = {⟨3⟩, 1 + ⟨3⟩, 2 + ⟨3⟩} ≅ ℤ₃, where ⟨3⟩ = {0, 3, 6, 9}.

ℤ₁₂/⟨4⟩ = {⟨4⟩, 1 + ⟨4⟩, 2 + ⟨4⟩, 3 + ⟨4⟩} ≅ ℤ₄, where ⟨4⟩ = {0, 4, 8}.

ℤ₁₂/⟨6⟩ = {k + ⟨6⟩ : k = 0, 1, ..., 5} ≅ ℤ₆, where ⟨6⟩ = {0, 6}.

Answer: up to isomorphism the quotient groups of ℤ₁₂ are ℤ₁₂, ℤ₆, ℤ₄, ℤ₃, ℤ₂ and the trivial group, all cyclic.

Example 11 (UPSC 2020, 10 marks): Let S₃ and Z₃ be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S₃ in Z₃ except the trivial homomorphism.

Let f : S₃ → ℤ₃ be a homomorphism. A transposition τ has order 2, so o(f(τ)) divides 2; it also divides o(ℤ₃) = 3. Hence o(f(τ)) = 1, i.e. f(τ) = 0.

S₃ is generated by its transpositions (for example (1 2 3) = (1 3)(1 2)), so f(σ) = 0 for every σ ∈ S₃.

Alternatively: a non-trivial f would have image of order 3, so ker f would be a normal subgroup of order 6/3 = 2. But no subgroup of order 2 is normal; e.g. (1 3)(1 2)(1 3) = (2 3) ∉ {e, (1 2)}.

Answer: the only homomorphism S₃ → ℤ₃ is the trivial one.

Example 12 (UPSC 2023, 10 marks): Let G be a group of order 10 and G′ be a group of order 6. Examine whether there exists a homomorphism of G onto G′.

Suppose f : G → G′ is an onto homomorphism. By the first isomorphism theorem G/ker f ≅ G′, so o(G′) = o(G)/o(ker f).

Then 6 · o(ker f) = 10, which is impossible since 6 does not divide 10.

(Homomorphisms into G′ do exist, e.g. the trivial one; every image has order dividing gcd(10, 6) = 2.)

Answer: no homomorphism of G onto G′ exists.

Example 13 (UPSC 2024, 15 marks): Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Let G be abelian and f : G → G′ a homomorphism. For x, y ∈ f(G) write x = f(a), y = f(b). Then xy = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = yx.

So f(G) is abelian; equivalently, every quotient of an abelian group is abelian.

The converse fails: sgn : S₃ → {1, −1} is an onto homomorphism with abelian image, but S₃ is not abelian, since (1 2)(1 3) = (1 3 2) while (1 3)(1 2) = (1 2 3).

Answer: homomorphic images of abelian groups are abelian; S₃ → {±1} shows that a non-abelian group can have an abelian image.

17. Practice Exercises

Exercise 1 (basic): Is V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} a normal subgroup of S₄?

Answer: yes. V is closed (the product of two of the non-identity elements is the third), and it is the union of the classes {e} and 'all products of two disjoint transpositions', so every conjugate stays in V.

Exercise 2 (basic): Is the subgroup D of diagonal matrices normal in GL(2, ℝ)?

Answer: no. With g = [[1, 1], [0, 1]] and d = diag(1, 2), gdg⁻¹ = [[1, 1], [0, 2]], which is not diagonal.

Exercise 3 (basic): Find the order of the element 3 + ⟨6⟩ in ℤ₁₈/⟨6⟩, and identify the quotient group.

Answer: ⟨6⟩ = {0, 6, 12}, so the quotient has order 6 and is ≅ ℤ₆. Since 3 ∉ ⟨6⟩ and 2·3 = 6 ∈ ⟨6⟩, the order is 2.

Exercise 4 (standard): Find all homomorphisms from ℤ₈ to ℤ₁₂. Is any of them onto?

Answer: f(x) = ax with 8a ≡ 0 (mod 12), i.e. a ∈ {0, 3, 6, 9}: gcd(8, 12) = 4 homomorphisms. None is onto, since an image has at most 8 elements.

Exercise 5 (standard): In G = ℤ × ℤ let N = {(a, b) : a + b is even}. Show that N ⊴ G and G/N ≅ ℤ₂.

Answer: f(a, b) = (a + b) mod 2 is an onto homomorphism G → ℤ₂ with kernel N; apply the first isomorphism theorem (N is normal as a kernel, or because G is abelian).

Exercise 6 (standard): Show that ℂ^*/ℝ⁺ ≅ T, where ℝ⁺ is the group of positive reals and T the unit circle, all under multiplication.

Answer: f(z) = z/|z| satisfies f(zw) = f(z)f(w), is onto (f(u) = u for u ∈ T), and f(z) = 1 iff z is a positive real. So ker f = ℝ⁺.

Exercise 7 (standard): Let N ⊴ G with [G : N] = m finite. Prove that g^m ∈ N for every g ∈ G.

Answer: G/N is a group of order m, so by Lagrange (gN)^m = N, i.e. g^mN = N, i.e. g^m ∈ N.

Exercise 8 (UPSC level): Prove that a group G is abelian if and only if the map x ↦ x⁻¹ is an automorphism of G.

Answer: the map is always a bijection (it is its own inverse). It is a homomorphism iff (ab)⁻¹ = a⁻¹b⁻¹, i.e. b⁻¹a⁻¹ = a⁻¹b⁻¹, for all a, b; taking inverses, this says ab = ba.

Exercise 9 (UPSC level): H ≤ G and N ⊴ G are finite with o(H) = 10, o(N) = 15 and o(H ∩ N) = 5. Find o(HN) and [HN : N], and check the second isomorphism theorem.

Answer: o(HN) = 10·15/5 = 30 and [HN : N] = 30/15 = 2, which equals [H : H ∩ N] = 10/5 = 2, as HN/N ≅ H/(H ∩ N) requires.

Exercise 10 (UPSC level): Find all normal subgroups of S₃ and, up to isomorphism, all quotient groups of S₃.

Answer: {e}, A₃ and S₃ (the three subgroups of order 2 are not normal). The quotients are S₃/{e} ≅ S₃, S₃/A₃ ≅ ℤ₂ and the trivial group.

Exercise 11 (UPSC level): Show that ℚ/ℤ is an infinite group in which every element has finite order.

Answer: q·(p/q + ℤ) = p + ℤ = ℤ, so p/q + ℤ has order dividing q. The cosets 1/n + ℤ (n ≥ 2) are distinct, since 1/m − 1/n ∉ ℤ for 2 ≤ m < n, so ℚ/ℤ is infinite.

1. Understanding Permutations and Cycle Notation

Why we need it: rearrangements of objects appear everywhere: shuffling cards, relabelling the vertices of a polygon, permuting the roots of a polynomial. By Cayley's theorem every group lives among them.

To compute with them we need a short notation and a fixed rule for composing two rearrangements.

In words: a permutation σ of {1, 2, ..., n} is a bijection: every symbol goes to exactly one symbol, and no two symbols go to the same place. S_n is the set of all of them; it has n! elements.

The cycle (a₁ a₂ ... a_k) sends each a_i to the next one and a_k back to a₁; symbols not written are fixed. The product στ means 'first τ, then σ' (right to left), like composition of functions.

First example, slowly: in S₅ let σ(1) = 3, σ(2) = 5, σ(3) = 1, σ(4) = 4, σ(5) = 2.

Step 1: start at 1 and follow the arrows: 1 → 3 → 1. The cycle closes, giving (1 3).

Step 2: the smallest unused symbol is 2: 2 → 5 → 2, giving (2 5). Step 3: 4 → 4 is fixed and is not written. So σ = (1 3)(2 5).

Step 4: compose with τ = (1 2 3), applying τ first. 1: τ(1) = 2, σ(2) = 5, so 1 → 5. 5: τ(5) = 5, σ(5) = 2, so 5 → 2. 2: τ(2) = 3, σ(3) = 1, so 2 → 1. 3 and 4 are fixed. Hence στ = (1 5 2).

Step 5: in the other order, τσ = (2 5 3). So στ ≠ τσ: notice that the order of composition matters in S_n.

Non-example: the rule 1 → 2, 2 → 2, 3 → 1 on {1, 2, 3} is not a permutation, because 1 and 2 both go to 2 (it is not one-one). Also (1 2)(2 3) is a product of cycles but not of disjoint cycles; its disjoint form is (1 2 3).

Building up: (1) Following arrows always closes a cycle, because σ is one-one and there are finitely many symbols. (2) Disjoint cycles commute, since they move different symbols.

(3) So the order of σ is the lcm of its disjoint cycle lengths: (1 3)(2 5) has order 2. (4) The inverse reverses each cycle: (1 5 2)⁻¹ = (2 5 1) = (1 2 5).

Where it is used: every computation in S_n (orders, powers, parity, conjugates) starts by writing disjoint cycles, and Cayley's theorem turns any finite group into such permutations.

2. Understanding Even and Odd Permutations

Why we need it: every rearrangement can be done by swapping two objects at a time. We want to know whether 'an even number of swaps' is a property of the permutation itself, not of the way we chose to do it.

The answer gives the alternating group A_n and the sign that appears in every determinant.

In words: a transposition (a b) swaps two symbols and fixes the rest. Every permutation is a product of transpositions in many ways, but by the parity theorem the number of factors is always even or always odd.

σ is even or odd accordingly, and sgn σ = +1 or −1. A k-cycle is a product of k − 1 transpositions, so its sign is (−1)^k−1.

First example, slowly: take the 5-cycle σ = (1 4 2 5 3). The rule (a₁ ... a_k) = (a₁ a_k)(a₁ a_k−1)···(a₁ a₂) gives σ = (1 3)(1 5)(1 2)(1 4).

Step 1 (check one symbol): applying the factors right to left, 1 → 4 under (1 4), and the other three factors fix 4; so 1 → 4, as in σ.

Step 2: σ is a product of 4 transpositions, so σ is even. Inserting (1 2)(1 2) = e gives a product of 6 transpositions: still even, as the parity theorem predicts.

Step 3 (a product of cycles): for τ = (1 2)(3 4 5 6), add (k − 1) over the disjoint cycles: 1 + 3 = 4, so τ is even.

Non-example: parity cannot be read off from the order. (1 2) and (1 2)(3 4) both have order 2, but the first is odd and the second is even.

Building up: (1) The parity theorem is proved with the product Δ (Derivation 'Why Parity is Well Defined' below). (2) Hence sgn : S_n → {1, −1} is a homomorphism, onto for n ≥ 2.

(3) A_n = ker sgn is normal of index 2 and order n!/2 (Standard Proofs, Theorem 8). (4) The odd permutations form the coset (1 2)A_n; they are not a subgroup, because they do not contain e.

Where it is used: deciding membership in A_n, the sign in the determinant expansion, showing that a subgroup of odd order lies in A_n, and the classic 15-puzzle impossibility.

3. Understanding Conjugacy and Cycle Type in S_n

Why we need it: two permutations may do 'the same thing' to differently named symbols. Recognising this answers three common questions: are σ and ρ conjugate, how many elements of a given order are there, and which subgroups are normal?

In words: τστ⁻¹ is σ with every symbol renamed by τ: replace each symbol i in the cycle form of σ by τ(i). The cycle type of σ is the list of its disjoint cycle lengths, for example (3)(2) for (1 2 3)(4 5).

First example, slowly: let σ = (1 2 3)(4 5) and τ = (1 5 3) in S₅, so τ(1) = 5, τ(2) = 2, τ(3) = 1, τ(4) = 4, τ(5) = 3.

Step 1: rename inside each cycle: (1 2 3) becomes (τ(1) τ(2) τ(3)) = (5 2 1), and (4 5) becomes (τ(4) τ(5)) = (4 3).

Step 2: so τστ⁻¹ = (5 2 1)(4 3) = (1 5 2)(3 4). Step 3 (check one symbol directly): τ⁻¹(5) = 1, σ(1) = 2, τ(2) = 2, so 5 → 2, as the cycle (1 5 2) says.

What to notice: the cycle lengths 3 and 2 did not change; only the names inside the cycles did.

Non-example: (1 2 3) and (1 2)(3 4) in S₄ are not conjugate, because their cycle types (3)(1) and (2)(2) differ; no renaming can turn a 3-cycle into two 2-cycles. Their orders, 3 and 2, differ as well.

Building up: (1) The relabelling rule is proved in the derivation 'Relabelling Rule and Class Sizes' below. (2) Conversely, if σ and ρ have the same cycle type, write one above the other with cycles of equal length aligned; reading down the columns gives τ with τστ⁻¹ = ρ.

(3) So the conjugacy classes of S_n are exactly the cycle types, with sizes n!/∏(k^m_k·m_k!). (4) A normal subgroup is a union of whole classes, which is why V₄ ⊴ S₄.

Where it is used: counting elements of a given order, writing class equations, finding τ with τστ⁻¹ = ρ, and proving that A₄ has no subgroup of order 6.

4. Understanding Cayley's Theorem

Why we need it: is the study of permutation groups a special case, or does every group appear there? Cayley's theorem says every group is isomorphic to a group of permutations, so nothing is lost by thinking of group elements as rearrangements.

In words: for each a ∈ G, the map T_a(x) = ax rearranges the set G (it is a permutation of G). The rule a ↦ T_a is a one-one homomorphism from G into Sym(G).

'Isomorphic to a subgroup of Sym(G)' means: the image {T_a : a ∈ G} is a subgroup of Sym(G), and G is isomorphic to that image.

First example, slowly: take G = ℤ₃ = {0, 1, 2} under addition, and regard 0, 1, 2 as the symbols being moved.

Step 1: T₀(x) = x is the identity permutation. Step 2: T₁ sends 0 → 1, 1 → 2, 2 → 0, which is the cycle (0 1 2).

Step 3: T₂ sends 0 → 2, 2 → 1, 1 → 0, which is (0 2 1). Step 4 (homomorphism check): T₁T₁ = (0 1 2)² = (0 2 1) = T₂ = T₁₊₁.

Step 5: the image {e, (0 1 2), (0 2 1)} is the alternating group on the three symbols, a copy of ℤ₃ inside S₃. Notice that each T_a is just the row of a in the Cayley table.

Non-example: right multiplication R_a(x) = xa also gives permutations, but a ↦ R_a reverses products: R_aR_b(x) = xba = R_ba(x). In a non-abelian group this is not a homomorphism; the repair is a ↦ R_a⁻¹.

Building up: (1) T_a is one-one by cancellation and onto because T_a(a⁻¹y) = y. (2) T_aT_b = T_ab by associativity. (3) T_a = identity forces a = ae = e. (4) Conclude by the first isomorphism theorem (Standard Proofs, Theorem 7).

Acting on the cosets of a subgroup H instead of on G gives the generalised Cayley theorem stated below, which embeds G/K in a much smaller S_n.

Where it is used: the standard proof question itself, embedding a group of order n in S_n, and index arguments such as 'A₅ has no subgroup of index 2, 3 or 4'.

5. Permutations and Cycles

Permutation: a bijection of a set onto itself. S_n is the group of all permutations of {1, 2, ..., n} under composition, and o(S_n) = n!.

Idea: think of a permutation as a rearrangement of labelled seats; composing two permutations means doing one rearrangement after the other. The order of doing them matters, which is why the Cayley table of S₃ in the figure above is not symmetric about its main diagonal.

Non-abelian: S_n is non-abelian for n ≥ 3; for example (1 2)(1 3) ≠ (1 3)(1 2).

Cycle: (a₁ a₂ ... a_k) sends a₁ → a₂ → ... → a_k → a₁ and fixes all other symbols; a 2-cycle is called a transposition.

Disjoint cycle decomposition: every permutation is a product of disjoint cycles, unique apart from the order of the factors; disjoint cycles commute.

Why it works: start at the smallest symbol a and follow a → σ(a) → σ²(a) → ...; since there are finitely many symbols and σ is one-one, the first repeated value is a itself, which closes a cycle. Repeat with the smallest symbol not yet used. Disjoint cycles commute because they move different symbols.

Order: the order of a permutation is the lcm of the lengths of its disjoint cycles.

Why it works: if σ = c₁c₂···c_r with disjoint cycles of lengths k_i, then σ^m = c₁^m···c_r^m (they commute), and this is the identity iff each c_i^m = e, i.e. iff every k_i divides m. The rule needs disjointness: (1 2)(2 3) = (1 2 3) has order 3, not lcm(2, 2) = 2.

Inverse: (a₁ a₂ ... a_k)⁻¹ = (a_k ... a₂ a₁).

Convention: in these notes στ means apply τ first, then σ (right to left); state your convention in the answer.

6. Even and Odd Permutations

Transpositions: (a₁ a₂ ... a_k) = (a₁ a_k)(a₁ a_k−1)...(a₁ a₂), so a k-cycle is a product of k − 1 transpositions.

Parity theorem: no permutation can be written both as a product of an even number and of an odd number of transpositions.

Sign rule: if the disjoint cycles have lengths k₁, ..., k_r, the permutation is even iff ∑(k_i − 1) is even; a cycle of even length is odd.

Alternating group: A_n, the set of even permutations, is a normal subgroup of S_n of index 2, so o(A_n) = n!/2 for n ≥ 2.

Idea: A_n is the kernel of the sign homomorphism sgn : S_n → {1, −1}, which is onto for n ≥ 2; so A_n is normal of index 2 automatically (proved in Standard Proofs, Theorem 8). The odd permutations form the other coset (1 2)A_n and do not form a subgroup.

Generators: S_n is generated by the transpositions, and also by the two elements (1 2) and (1 2 ... n); A_n (n ≥ 3) is generated by the 3-cycles.

Why it works: every cycle is a product of transpositions, so the transpositions generate S_n. An even permutation is a product of pairs of transpositions, and each pair is a product of 3-cycles: (a b)(a c) = (a c b), and for distinct a, b, c, d, (a b)(c d) = (a c b)(a c d). So the 3-cycles generate A_n.

7. Conjugacy in S_n

Relabelling rule: σ(a₁ a₂ ... a_k)σ⁻¹ = (σ(a₁) σ(a₂) ... σ(a_k)).

Conjugacy classes: two permutations are conjugate in S_n iff they have the same cycle type.

Idea: conjugating by σ only renames the symbols, so it cannot change the lengths of the cycles; conversely two permutations with the same cycle type can be matched cycle by cycle, and the matching is the renaming σ (Example 6 below).

Counting: the number of k-cycles in S_n is n!/((n − k)!·k).

Check in S₄: identity 1, transpositions 6, 3-cycles 8, 4-cycles 6, products of two disjoint transpositions 3; total 24.

8. Cayley's Theorem

Statement: every group G is isomorphic to a subgroup of Sym(G), the group of all permutations of the set G. In particular, a group of order n is isomorphic to a subgroup of S_n.

Idea: every group is a group of symmetries, namely of its own underlying set: the element a 'shuffles' G by x ↦ ax. In the Cayley table (figure above) the row of a is exactly this shuffle, and each row is a permutation of the elements because of the cancellation law. The proof below is also given in Standard Proofs, Theorem 7.

Proof:

1. For a ∈ G define T_a : G → G by T_a(x) = ax.

2. T_a is one-one, since ax = ay ⇒ x = y, and onto, since T_a(a⁻¹y) = y; so T_a ∈ Sym(G).

3. (T_aT_b)(x) = a(bx) = (ab)x = T_ab(x), so φ(a) = T_a is a homomorphism from G to Sym(G).

4. If T_a is the identity map then ae = e, so a = e; hence φ is one-one.

5. Therefore G ≅ φ(G) = {T_a : a ∈ G}, a subgroup of Sym(G).

Generalised Cayley theorem: if H ≤ G with [G : H] = n, then G acting on the left cosets of H gives a homomorphism G → S_n whose kernel is contained in H.

Use: the kernel K is a normal subgroup of G inside H, and G/K embeds in S_n, so [G : K] divides n!. For n = 2 this forces K = H, which re-proves 'index 2 ⇒ normal'. For a simple group K = {e}, so |G| divides n!; hence A₅ (order 60) has no subgroup of index 2, 3 or 4, i.e. none of order 30, 20 or 15.

Example 1: Find the order and parity of σ = (1 2 3)(4 5) in S₅.

The cycles are disjoint with lengths 3 and 2, so o(σ) = lcm(3, 2) = 6. σ is a product of 2 + 1 = 3 transpositions, so σ is odd.

Example 2: Compute (1 2 3)(2 3 4) in S₄ (right to left).

1 → 1 → 2; 2 → 3 → 1; 3 → 4 → 4; 4 → 2 → 3. Hence (1 2 3)(2 3 4) = (1 2)(3 4), an even permutation, as expected for a product of two 3-cycles.

9. Derivation: Why Parity is Well Defined

Step 1: let Δ = ∏_{i < j} (x_j − x_i), the product over all pairs 1 ≤ i < j ≤ n. For σ ∈ S_n let σΔ = ∏_{i < j} (x_σ(j) − x_σ(i)). Each pair {i, j} still appears exactly once, possibly reversed, so σΔ = ±Δ; write σΔ = sgn(σ)Δ.

Step 2: applying τ and then σ to the variables gives (στ)Δ = σ(τΔ) = sgn(τ)·σΔ = sgn(σ)sgn(τ)Δ, so sgn is a homomorphism S_n → {1, −1}.

Step 3: for a transposition (a b) with a < b, the factor x_b − x_a changes sign. For each other symbol c, the two factors involving c and a or b are exchanged with each other, and their sign changes cancel in pairs. So sgn(a b) = −1.

Conclusion: a product of r transpositions has sign (−1)^r. If σ were a product of both an even and an odd number of transpositions, sgn σ would equal both 1 and −1. Hence parity is well defined, and sgn of a k-cycle is (−1)^k−1.

10. Derivation: Relabelling Rule and Class Sizes

Relabelling rule: let c = (a₁ ... a_k). Then σcσ⁻¹(σ(a_i)) = σ(c(a_i)) = σ(a_i+1) (indices mod k). If x is not one of the σ(a_i), then σ⁻¹(x) is fixed by c, so σcσ⁻¹(x) = x. Hence σcσ⁻¹ = (σ(a₁) ... σ(a_k)); apply this to each disjoint cycle.

Counting k-cycles: an ordered list of k distinct symbols can be chosen in n!/(n − k)! ways, and each k-cycle is written in exactly k ways (start at any of its symbols). So there are n!/((n − k)!·k) k-cycles; for 3-cycles in S₄, 24/(1·3) = 8.

General class size: if the cycle type has m_k cycles of length k (fixed points counted as 1-cycles, ∑k·m_k = n), the number of such permutations is n!/∏_k(k^m_k·m_k!). The extra m_k! is there because cycles of equal length can be listed in any order. Check: type (2)(2) in S₄ gives 24/(2²·2!) = 3.

Normal subgroups of S_n: a normal subgroup is closed under conjugation, so it is a union of whole conjugacy classes. In S₄, V₄ = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is the union of the classes of sizes 1 and 3, hence V₄ ⊴ S₄.

11. Choosing the Method

If the order of a permutation is asked → first write it as disjoint cycles, then take the lcm of the lengths.

If parity is asked → compute ∑(k_i − 1) over the disjoint cycles; do not try to guess it from the order.

If a high power σ^N is asked → reduce N modulo each cycle length separately.

If you must find τ with τστ⁻¹ = ρ → write σ and ρ in cycle form with matching cycle lengths one under the other and read τ symbol by symbol.

If you must count elements of a given order in S_n → list the cycle types whose lcm is that order and add their class sizes.

If you must decide whether H ≤ S_n is normal → check whether H is a union of conjugacy classes (same cycle types throughout).

If you must embed a group in a permutation group → use left multiplication on G (Cayley), or on the cosets of a subgroup for a smaller S_n.

12. Common Mistakes

Mixing conventions: computing στ left to right in one line and right to left in the next. State the convention once and keep it.

Taking the lcm of overlapping cycles: the lcm rule is for disjoint cycles only; (1 2)(2 3) = (1 2 3) has order 3.

Getting the parity of a cycle backwards: a cycle of even length is odd, and a cycle of odd length is even, because a k-cycle is k − 1 transpositions.

Over-counting cycles: forgetting to divide by k for the k ways of writing a k-cycle, or by m_k! for cycles of equal length.

Using S_n-classes inside A_n: a class of S_n may split in A_n; the eight 3-cycles of A₄ form two classes of size 4, so the class equation of A₄ is 12 = 1 + 3 + 4 + 4.

13. Connections

Link: the sign of a permutation is the sign in the expansion det A = ∑_σ sgn(σ) a_1σ(1)a_2σ(2)···a_nσ(n) of Linear Algebra; this is why interchanging two rows changes the sign of a determinant.

Link: sending σ to its permutation matrix embeds S_n in GL(n, ℝ) with det = sgn σ; combined with Cayley's theorem, every finite group is a group of matrices.

14. Worked Examples

Example 3: Find the number of elements of order 6 in S₆.

The order is the lcm of the disjoint cycle lengths, and the lengths add up to at most 6. lcm = 6 is possible only for a 6-cycle, or for a 3-cycle times a disjoint 2-cycle (one point fixed).

Number of 6-cycles = 6!/6 = 120.

Type (3)(2): choose and arrange the 3-cycle in C(6, 3)·2 = 40 ways and the 2-cycle from the remaining 3 points in 3 ways, giving 120.

Total = 120 + 120 = 240.

Example 4: Show that A₄ has no subgroup of order 6.

Suppose H ≤ A₄ with o(H) = 6. Then [A₄ : H] = 2, so H ⊴ A₄ and A₄/H has order 2; hence g² ∈ H for every g ∈ A₄.

If σ is a 3-cycle, then σ³ = e, so σ = σ⁴ = (σ²)² ∈ H. Thus all eight 3-cycles of A₄ lie in H.

Together with e, H has at least 9 elements, contradicting o(H) = 6. So A₄ has no subgroup of order 6, and the converse of Lagrange's theorem is false.

Example 5: Let σ ∈ S₆ be given by σ(1) = 2, σ(2) = 4, σ(3) = 5, σ(4) = 1, σ(5) = 3, σ(6) = 6. Write σ as a product of disjoint cycles, find its order and parity, and compute σ¹⁰⁰.

Following 1 → 2 → 4 → 1 and 3 → 5 → 3 gives σ = (1 2 4)(3 5), with 6 fixed.

o(σ) = lcm(3, 2) = 6. σ is a product of 2 + 1 = 3 transpositions, so σ is odd.

Disjoint cycles commute, so σ¹⁰⁰ = (1 2 4)¹⁰⁰(3 5)¹⁰⁰. As 100 ≡ 1 (mod 3) and 100 ≡ 0 (mod 2), σ¹⁰⁰ = (1 2 4).

Example 6: Find τ ∈ S₅ such that τ(1 2 3)(4 5)τ⁻¹ = (2 4 5)(1 3).

By the relabelling rule, τ(1 2 3)(4 5)τ⁻¹ = (τ(1) τ(2) τ(3))(τ(4) τ(5)). Match this with (2 4 5)(1 3).

Take τ(1) = 2, τ(2) = 4, τ(3) = 5, τ(4) = 1, τ(5) = 3, i.e. τ = (1 2 4)(3 5).

The answer is not unique: writing the target as (4 5 2)(3 1), for example, gives another valid τ. Such τ exists because both permutations have cycle type (3)(2).

15. Practice Exercises

Exercise 1 (basic): In S₆ let σ(1) = 4, σ(2) = 6, σ(3) = 1, σ(4) = 3, σ(5) = 2, σ(6) = 5. Write σ in disjoint cycles and find its order and parity.

Answer: σ = (1 4 3)(2 6 5); order lcm(3, 3) = 3; even, since 2 + 2 = 4 transpositions.

Exercise 2 (basic): Compute (1 3 5)(1 2) and (1 2)(1 3 5) in S₅, composing right to left.

Answer: (1 3 5)(1 2) = (1 2 3 5) and (1 2)(1 3 5) = (1 3 5 2); they differ.

Exercise 3 (basic): Write the inverse of (1 2 4 3)(5 6).

Answer: (3 4 2 1)(6 5) = (1 3 4 2)(5 6).

Exercise 4 (standard): How many elements of order 2 are there in S₅?

Answer: 25: transpositions C(5, 2) = 10, and products of two disjoint transpositions 5·3 = 15.

Exercise 5 (standard): For σ = (1 2 3 4 5)(6 7 8) in S₈, find σ⁹⁹.

Answer: 99 ≡ 4 (mod 5) and 99 ≡ 0 (mod 3), so σ⁹⁹ = (1 2 3 4 5)⁴ = (1 2 3 4 5)⁻¹ = (1 5 4 3 2).

Exercise 6 (standard): Does (1 2 3 4)(5 6 7) belong to A₇? Find its order.

Answer: no: it is a product of 3 + 2 = 5 transpositions, hence odd. Its order is lcm(4, 3) = 12.

Exercise 7 (standard): Find τ ∈ S₅ with τ(1 2)(3 4 5)τ⁻¹ = (2 5)(1 3 4).

Answer: match symbol by symbol: τ(1) = 2, τ(2) = 5, τ(3) = 1, τ(4) = 3, τ(5) = 4, i.e. τ = (1 2 5 4 3) (other choices also work).

Exercise 8 (UPSC level): Find the largest order of an element of S₇ and give an element of that order.

Answer: 12, from cycle type (4)(3), e.g. (1 2 3)(4 5 6 7). The other types give at most lcm(5, 2) = 10, 7, 6 or less.

Exercise 9 (UPSC level): Prove that S_n (n ≥ 2) is generated by the transpositions (1 2), (1 3), ..., (1 n).

Answer: every permutation is a product of transpositions, and for a, b ≠ 1, (a b) = (1 a)(1 b)(1 a).

Exercise 10 (UPSC level): Show that V₄ = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is a normal subgroup of A₄, and identify A₄/V₄.

Answer: V₄ is a union of conjugacy classes of S₄, so it is normal in S₄ and hence in A₄. A₄/V₄ has order 12/4 = 3, so it is ≅ ℤ₃.

Exercise 11 (UPSC level): How many conjugacy classes does S₅ have? Give the size of each and check that they add up to 120.

Answer: 7, one per partition of 5: e (1), (2) (10), (3) (20), (4) (30), (5) (24), (2)(2) (15), (3)(2) (20); 1 + 10 + 20 + 30 + 24 + 15 + 20 = 120.

1. Understanding Direct Products

Why we need it: we want to build larger groups from smaller ones, and, in the other direction, to recognise a given group as made of simpler pieces, for instance ℤ₆ as ℤ₂ and ℤ₃ working side by side.

In words: G₁ × G₂ is the set of ordered pairs (g₁, g₂). Pairs are combined coordinate by coordinate, each coordinate using the operation of its own group, so its order is o(G₁)·o(G₂).

A power of a pair is the pair of powers, so a pair returns to the identity only when both coordinates have returned. That is why o((g₁, g₂)) is the lcm of the two orders.

First example, slowly: in ℤ₂ × ℤ₃ list the multiples of (1, 1): (1, 1), (0, 2), (1, 0), (0, 1), (1, 2), (0, 0).

Step 1: the first coordinate cycles with period 2 and the second with period 3. Step 2: both are 0 for the first time at the 6th step, and 6 = lcm(2, 3).

Step 3: all six elements appeared, so (1, 1) generates the group, and ℤ₂ × ℤ₃ ≅ ℤ₆.

Non-example: in ℤ₂ × ℤ₂, every element added to itself gives (0, 0), so every element has order 1 or 2. No element has order 4, so ℤ₂ × ℤ₂ is not ≅ ℤ₄; here gcd(2, 2) ≠ 1.

Building up: (1) o((g₁, g₂)) = lcm(o(g₁), o(g₂)). (2) Hence ℤ_m × ℤ_n is cyclic iff gcd(m, n) = 1 (the 'Why it works' line below).

(3) Inside one group, G ≅ H × K when H, K are normal, H ∩ K = {e} and G = HK. (4) Every finite abelian group is a product of cyclic groups of prime-power order, so its structure reduces to partitions of exponents.

Where it is used: listing abelian groups of order n, the Chinese remainder theorem ℤ_mn ≅ ℤ_m × ℤ_n, U(mn) ≅ U(m) × U(n), and groups of order pq.

2. Understanding Dihedral Groups

Why we need it: the symmetries of a regular polygon form the most important family of non-abelian groups. They are small enough to compute by hand, and they supply the standard counter-examples of this chapter.

In words: r is the rotation through 2π/n and s is one reflection. The relations rⁿ = e, s² = e and srs⁻¹ = r⁻¹ say: n turns bring you back, two flips bring you back, and a turn seen in a mirror is a turn the other way.

Every element is then a rotation r^k or a reflection sr^k (0 ≤ k ≤ n − 1), so o(D_n) = 2n.

First example, slowly: label the vertices of a square 1, 2, 3, 4 anticlockwise. Then r = (1 2 3 4), and the reflection in the diagonal through vertices 1 and 3 is s = (2 4).

Step 1: by the relabelling rule, srs⁻¹ = (s(1) s(2) s(3) s(4)) = (1 4 3 2) = r⁻¹. The relation holds.

Step 2: sr = (2 4)(1 2 3 4): 1 → 2 → 4, 4 → 1 → 1, 2 → 3 → 3, 3 → 4 → 2. So sr = (1 4)(2 3), the reflection in the line through the mid-points of sides 14 and 23. It has order 2, like every reflection.

Step 3: rs = (1 2 3 4)(2 4) = (1 2)(3 4) ≠ sr, so D₄ is not abelian.

Non-example: Q₈ is also a non-abelian group of order 8, but it is not ≅ D₄. D₄ has five elements of order 2 (four reflections and r²), while −1 is the only element of order 2 in Q₈.

Building up: (1) From srs⁻¹ = r⁻¹ we get sr = r⁻¹s, and by induction sr^k = r^−ks: a reflection can be moved past a rotation by inverting the rotation.

(2) Every reflection has order 2: (sr^k)² = s(r^ks)r^k = s(sr^−k)r^k = e. (3) The centre is {e} for odd n and {e, r^n/2} for even n (the 'Why it works' line below). (4) ⟨r⟩ has index 2, hence is normal.

Where it is used: groups of order 2p, groups generated by two elements of order 2, the non-transitivity of normality, and class-equation computations.

3. Understanding Conjugacy Classes and the Class Equation

Why we need it: sorting the elements of G into conjugacy classes shows at once which elements are central, and the sizes of the classes obey strong divisibility rules. These rules prove, for example, that a group of order pⁿ has a non-trivial centre.

In words: cl(a) = {gag⁻¹ : g ∈ G} collects every element that 'looks like a' from some viewpoint. C(a) = {g : ga = ag} is the set of elements that do not move a under conjugation.

The more elements commute with a, the smaller its class: |cl(a)| = [G : C(a)]. The class equation adds the class sizes, listing the one-element classes (the centre) separately.

First example, slowly: find the classes of D₄ = {e, r, r², r³, s, sr, sr², sr³}, using sr^k = r^−ks.

Step 1: Z(D₄) = {e, r²}, giving two classes of size 1. Step 2: srs⁻¹ = r⁻¹ = r³, so cl(r) = {r, r³}.

Step 3: rsr⁻¹ = r(sr⁻¹) = r(rs) = r²s = sr⁻² = sr², so cl(s) = {s, sr²}. Step 4: r(sr)r⁻¹ = rs = sr⁻¹ = sr³, so cl(sr) = {sr, sr³}.

Step 5: the class equation is 8 = 2 + 2 + 2 + 2, where the first 2 is o(Z(D₄)). Every class size divides 8, as it must.

Non-example: 8 = 1 + 1 + 3 + 3 cannot be the class equation of any group of order 8, because a class of size 3 would need 3 to divide 8.

Building up: (1) Conjugacy is an equivalence relation, so the classes partition G. (2) gC(a) ↦ gag⁻¹ is a bijection from the cosets of C(a) onto cl(a), which gives |cl(a)| = [G : C(a)].

(3) If o(G) = pⁿ, every non-central class size is a power of p bigger than 1, so p divides o(Z(G)) and Z(G) ≠ {e}. (4) With the G/Z(G) theorem this makes every group of order p² abelian (the derivation below).

Where it is used: centres of p-groups, groups of order p², class equations of S_n, A₄ and D_n, and proofs of Cauchy's and Sylow's theorems.

4. Understanding the Sylow Theorems

Why we need it: the converse of Lagrange's theorem is false, so a divisor of o(G) need not be the order of a subgroup. The Sylow theorems guarantee subgroups of the largest prime-power orders and count them, which often forces a normal subgroup.

In words: write o(G) = p^km with p prime and p not dividing m. A Sylow p-subgroup is a subgroup of order exactly p^k, the full power of p in o(G).

The theorems say: such subgroups exist; any two are conjugate; and their number n_p satisfies n_p ≡ 1 (mod p) and n_p | m. If n_p = 1, the Sylow p-subgroup equals all its conjugates, i.e. it is normal.

First example, slowly: let o(G) = 45 = 3²·5.

Step 1 (p = 3): n₃ divides 5, so n₃ ∈ {1, 5}; n₃ ≡ 1 (mod 3) rules out 5 (5 ≡ 2). So n₃ = 1, and the Sylow 3-subgroup P (order 9) is normal.

Step 2 (p = 5): n₅ divides 9, so n₅ ∈ {1, 3, 9}; modulo 5 these are 1, 3, 4, so n₅ = 1 and the Sylow 5-subgroup Q (order 5) is normal.

Step 3: P ∩ Q = {e}, since its order divides both 9 and 5, and |PQ| = 9·5 = 45. So G = PQ ≅ P × Q.

Step 4: P has order 3², hence is abelian, and Q ≅ ℤ₅. So every group of order 45 is abelian: it is ℤ₉ × ℤ₅ ≅ ℤ₄₅ or ℤ₃ × ℤ₃ × ℤ₅.

Non-example: for o(G) = 12 = 2²·3 the conditions allow n₃ ∈ {1, 4}, and A₄ really has n₃ = 4 (its eight 3-cycles lie in four subgroups of order 3). So the Sylow conditions alone do not always produce a normal subgroup; then element counting is needed.

Building up: (1) The conjugates of a Sylow subgroup P are all the Sylow p-subgroups, so n_p = [G : N(P)], which divides [G : P] = m. (2) n_p = 1 ⇔ P ⊴ G.

(3) Subgroups of prime order p meet only in e, so n_p of them contain n_p(p − 1) elements of order p; this is the counting used for order 12 below. (4) The full proofs of Sylow's theorems are beyond the syllabus; Cauchy's theorem is proved in the derivation below.

Where it is used: proving that groups of order 15, 33, 35, 45 or pq are cyclic or abelian, showing that groups of order 20, 30 or 63 are not simple, and counting elements of prime order.

5. Direct Products

External direct product: for groups G₁, ..., G_n, the set G₁ × ... × G_n = {(g₁, ..., g_n)} with componentwise operation is a group of order o(G₁)···o(G_n); it is abelian iff every G_i is abelian.

Idea: picture independent clocks running side by side: an element (g₁, g₂) records the position of each, and the operation moves each clock by its own rule. Nothing in one coordinate affects the other.

Order of an element: o((g₁, ..., g_n)) = lcm(o(g₁), ..., o(g_n)).

Why it works: (g₁, ..., g_n)^m = (g₁^m, ..., g_n^m) is the identity iff g_i^m = e for every i, i.e. iff every o(g_i) divides m; the least such m is the lcm.

Cyclic products: ℤ_m × ℤ_n is cyclic, and then ≅ ℤ_mn, iff gcd(m, n) = 1. So ℤ₂ × ℤ₃ ≅ ℤ₆ but ℤ₂ × ℤ₂ ≇ ℤ₄; for rings, ℤ_mn ≅ ℤ_m × ℤ_n when gcd(m, n) = 1 (Chinese remainder theorem).

Why it works: if gcd(m, n) = 1, then o((1, 1)) = lcm(m, n) = mn, so (1, 1) generates. If d = gcd(m, n) > 1, every element has order dividing lcm(m, n) = mn/d < mn, so no element generates.

Internal direct product: if H, K ⊴ G, H ∩ K = {e} and G = HK, then G ≅ H × K via (h, k) ↦ hk; in that case hk = kh for all h ∈ H, k ∈ K.

Why it works: for h ∈ H and k ∈ K, the commutator hkh⁻¹k⁻¹ = (hkh⁻¹)k⁻¹ lies in K and = h(kh⁻¹k⁻¹) lies in H, since both are normal; so it is in H ∩ K = {e} and hk = kh. Then (h, k) ↦ hk is a homomorphism, one-one because hk = e ⇒ h = k⁻¹ ∈ H ∩ K, and onto because G = HK.

Both must be normal: S₃ = A₃·{e, (1 2)} with trivial intersection, but {e, (1 2)} is not normal, and S₃ is not ≅ ℤ₃ × ℤ₂ (which is abelian).

Fundamental theorem of finite abelian groups: every finite abelian group is a direct product of cyclic groups of prime-power order, uniquely up to order. For example, the abelian groups of order 8 are ℤ₈, ℤ₄ × ℤ₂ and ℤ₂ × ℤ₂ × ℤ₂.

How to list them: for n = p₁^a₁···p_r^a_r, choose for each prime a partition of its exponent a_i. So the number of abelian groups of order n is the product of the partition numbers; for 72 = 2³·3² it is 3·2 = 6.

Example 1: Is ℤ₂ × ℤ₄ cyclic?

The order of (a, b) is lcm(o(a), o(b)) with o(a) | 2 and o(b) | 4, so it is at most 4 < 8. No element has order 8, so ℤ₂ × ℤ₄ is not cyclic (consistent with gcd(2, 4) = 2 ≠ 1).

6. Dihedral Groups

Definition: D_n (n ≥ 3) is the group of symmetries of a regular n-gon: n rotations e, r, ..., rⁿ⁻¹ (r = rotation by 2π/n) and n reflections s, sr, ..., srⁿ⁻¹; o(D_n) = 2n.

Idea: r turns the polygon and s flips it over. Flipping reverses the sense of turning, which is exactly the relation srs⁻¹ = r⁻¹: flip, turn anticlockwise, flip back is the same as turning clockwise.

Presentation: D_n = ⟨r, s : rⁿ = s² = e, srs⁻¹ = r⁻¹⟩. Useful rules: sr^k = r^−ks, every reflection sr^k has order 2, and ⟨r⟩ is a normal subgroup of index 2.

Centre: Z(D_n) = {e} for odd n and {e, r^n/2} for even n. D₃ ≅ S₃, and D_n is non-abelian for n ≥ 3.

Why it works: r^k commutes with s iff r^−ks = r^ks iff r^2k = e iff n | 2k, so k = 0, or k = n/2 when n is even. A reflection sr^k never commutes with r for n ≥ 3, since r(sr^k) = sr^k−1 while (sr^k)r = sr^k+1.

Generated by two involutions: if u ≠ v have order 2 and uv has order n, then for n ≥ 3, ⟨u, v⟩ ≅ D_n (take r = uv, s = u; for n = 2 the group is the Klein four-group); conversely D_n is generated by the two reflections s and sr.

Groups of order 2p: for an odd prime p, a group of order 2p is either cyclic (ℤ_2p) or dihedral (D_p).

Example 2: List the elements of D₄ = ⟨r, s⟩ and find its centre.

D₄ = {e, r, r², r³, s, sr, sr², sr³}. r² commutes with s, because sr² = r⁻²s = r²s, and with every power of r.

r and r³ do not commute with s (sr = r³s ≠ rs), and no reflection sr^k commutes with r, since r(sr^k) = sr^k−1 ≠ sr^k+1. Hence Z(D₄) = {e, r²}.

7. Conjugacy, Class Equation and Cauchy's Theorem

Conjugacy: b is conjugate to a if b = gag⁻¹ for some g ∈ G; this is an equivalence relation whose classes cl(a) partition G, and cl(a) = {a} iff a ∈ Z(G).

Idea: conjugate elements do the same thing seen from a different viewpoint. In D_n, two reflections are conjugate when a symmetry carries one axis to the other: for odd n all n reflections form one class, and for even n they split into two classes of n/2 (axes through vertices and axes through mid-points of sides).

Class size: |cl(a)| = [G : C(a)], where C(a) is the centraliser of a; so the size of each class divides o(G).

Why it works: the map gC(a) ↦ gag⁻¹ from the left cosets of C(a) onto cl(a) is well defined and one-one, because gag⁻¹ = hah⁻¹ ⇔ h⁻¹g ∈ C(a) ⇔ gC(a) = hC(a). So |cl(a)| = [G : C(a)].

Class equation: o(G) = o(Z(G)) + ∑[G : C(a)], the sum running over one representative a of each class with more than one element.

Why it works: the conjugacy classes partition G, and the classes with one element are exactly {z} for z ∈ Z(G). Adding the class sizes gives the equation; every term divides o(G).

p-groups: if o(G) = pⁿ (p prime, n ≥ 1), then Z(G) ≠ {e}, since every term [G : C(a)] > 1 is a multiple of p. Every group of order p² is abelian (≅ ℤ_p² or ℤ_p × ℤ_p).

G/Z(G) theorem: if G/Z(G) is cyclic, then G is abelian. So o(G/Z(G)) can never be a prime.

Cauchy's theorem: if a prime p divides o(G), then G has an element of order p (and hence a subgroup of order p).

Normaliser: N(H) = {g ∈ G : gHg⁻¹ = H} is the largest subgroup of G in which H is normal; the number of conjugates of H is [G : N(H)].

Example 3: Write the class equation of S₃.

In S_n conjugacy classes are the cycle types, so the classes are {e}, the three transpositions and the two 3-cycles. Z(S₃) = {e}, so the class equation is 6 = 1 + 3 + 2; each class size divides 6.

8. Sylow Theorems and Groups of Small Order

Sylow theorems: let o(G) = p^km with p prime and p ∤ m. (1) G has a subgroup of order p^k (a Sylow p-subgroup). (2) Any two Sylow p-subgroups are conjugate. (3) The number n_p of Sylow p-subgroups satisfies n_p ≡ 1 (mod p) and n_p | m.

Idea: a Sylow p-subgroup is the largest possible p-piece of G. All of them are copies of one another (conjugate), and n_p = [G : N(P)], which is why n_p divides m. Counting n_p is the standard way to prove that a group of a given order has a normal subgroup.

Normality test: a Sylow p-subgroup is normal iff n_p = 1; more generally, a subgroup that is the only one of its order is normal.

Groups of order pq: let p < q be primes. The subgroup of order q is unique, hence normal; if moreover p ∤ (q − 1), the group is cyclic (so every group of order 15, 33 or 35 is cyclic).

Why it works: n_q ≡ 1 (mod q) and n_q | p < q force n_q = 1. Also n_p ∈ {1, q}, and n_p = q would need q ≡ 1 (mod p); if p ∤ (q − 1), then n_p = 1 too, both subgroups are normal with trivial intersection, and G ≅ ℤ_p × ℤ_q ≅ ℤ_pq.

Elementary method without Sylow: two distinct subgroups H, K of prime order m meet only in {e}, so |HK| = m²; if m² > o(G), there is at most one subgroup of order m.

Example 4: Show that every group of order 15 is cyclic.

n₅ ≡ 1 (mod 5) and n₅ | 3 give n₅ = 1; n₃ ≡ 1 (mod 3) and n₃ | 5 give n₃ = 1.

So G has unique, hence normal, subgroups H and K of orders 3 and 5. H ∩ K = {e} and |HK| = 15, so G = HK ≅ H × K ≅ ℤ₃ × ℤ₅ ≅ ℤ₁₅.

9. Derivation: Cauchy's Theorem (McKay's Counting Proof)

Setting: let p be a prime dividing o(G) and X = {(a₁, ..., a_p) ∈ G^p : a₁a₂···a_p = e}. The first p − 1 entries are arbitrary and then a_p is forced, so |X| = o(G)^p−1, a multiple of p.

Cyclic shift: if a₁a₂···a_p = e then a₂···a_pa₁ = a₁⁻¹(a₁···a_p)a₁ = e, so shifting the entries maps X to X. Repeating the shift p times returns the tuple, so each orbit has size 1 or p (a divisor of the prime p).

Counting: orbits of size 1 are the constant tuples (a, ..., a) with a^p = e. Since |X| and the sizes of the other orbits are multiples of p, the number of such a is a multiple of p. It is at least 1 (a = e), hence at least p, so some a ≠ e has a^p = e, i.e. o(a) = p.

10. Derivation: Using the Class Equation and Sylow Counting

Groups of order p² are abelian: Z(G) ≠ {e} (p-group), so o(G/Z(G)) is 1 or p. In both cases G/Z(G) is cyclic, so G is abelian (and then Z(G) = G).

Counting elements (order 12): n₃ ∈ {1, 4}. If n₃ = 4, the four subgroups of order 3 meet only in e and give 4·2 = 8 elements of order 3. Only 4 elements remain, and a Sylow 2-subgroup (order 4) contains no element of order 3, so it consists of exactly these 4 elements; it is unique, hence normal. So a group of order 12 has a normal Sylow 3-subgroup or a normal Sylow 2-subgroup.

Not simple: for o(G) = 20, n₅ ≡ 1 (mod 5) and n₅ | 4 give n₅ = 1, so the Sylow 5-subgroup is a proper non-trivial normal subgroup and G is not simple.

11. Choosing the Method

If asked whether G₁ × G₂ is cyclic → both factors must be cyclic of coprime orders; otherwise bound every element order by an lcm.

If asked to list abelian groups of order n → factor n and take partitions of each prime exponent.

If asked to show G = H × K internally → verify H ⊴ G, K ⊴ G, H ∩ K = {e} and |H||K| = |G|.

If you need an element of prime order p → Cauchy; a subgroup of order p^k → Sylow.

If you must show a group of order n is not simple → find p with n_p = 1; if every n_p could exceed 1, count the elements of prime order as above.

If the centre or abelianness of a p-group is asked → class equation, then the G/Z(G) theorem.

If conjugacy classes of D_n are asked → use sr^k = r^−ks: rotations pair up as {r^k, r^−k}, and reflections form one class (n odd) or two classes (n even).

12. Common Mistakes

Writing ℤ_m × ℤ_n ≅ ℤ_mn without gcd(m, n) = 1: ℤ₂ × ℤ₂ is the Klein group, not ℤ₄.

Taking o((a, b)) = o(a)·o(b): the order is the lcm; (1, 3) in ℤ₄ × ℤ₆ has order lcm(4, 2) = 4, not 4·2 = 8.

Misusing n_p: n_p divides m, the part of o(G) prime to p, not just o(G); and n_p = 1 means normal, not 'G is a p-group'.

Assuming distinct Sylow subgroups meet trivially: this is guaranteed only when they have prime order p; two Sylow 2-subgroups of order 4 can share an element of order 2.

Class-equation slips: putting central elements into the sum, or writing a class size that does not divide o(G).

13. Connections

Link: ℤ_mn ≅ ℤ_m × ℤ_n (gcd(m, n) = 1) is also a ring isomorphism (Rings topic); restricting it to units gives U(mn) ≅ U(m) × U(n), which is why φ(mn) = φ(m)φ(n).

Link: in Linear Algebra terms the rotations of D_n are orthogonal matrices with determinant 1 and the reflections have determinant −1, so ⟨r⟩ = D_n ∩ SO(2) is the kernel of det, again a normal subgroup of index 2.

14. Solved Past UPSC Questions

Example 5 (UPSC 2017, 15 marks): Show that the groups ℤ₅ × ℤ₇ and ℤ₃₅ are isomorphic.

ℤ₅ × ℤ₇ has order 35, and o((1, 1)) = lcm(5, 7) = 35.

So ℤ₅ × ℤ₇ = ⟨(1, 1)⟩ is cyclic of order 35, and every cyclic group of order 35 is isomorphic to ℤ₃₅.

Explicitly, φ(k) = (k mod 5, k mod 7) is a well-defined homomorphism ℤ₃₅ → ℤ₅ × ℤ₇ (5 and 7 divide 35). φ(k) = (0, 0) ⇔ 5 | k and 7 | k ⇔ 35 | k, so φ is one-one, hence onto, as both groups have 35 elements.

Answer: ℤ₅ × ℤ₇ ≅ ℤ₃₅, because gcd(5, 7) = 1.

Example 6 (UPSC 2023, 15 marks): Prove that a non-commutative group of order 2p, where p is an odd prime, must have a subgroup of order p.

Let o(G) = 2p. By Lagrange every element has order 1, 2, p or 2p.

No element has order 2p, otherwise G would be cyclic and hence commutative.

If every non-identity element had order 2, then ab = (ab)⁻¹ = b⁻¹a⁻¹ = ba for all a, b, and G would be abelian. This is excluded.

Hence some a ∈ G has order p, and H = ⟨a⟩ is a subgroup of order p (normal, since its index is 2).

Answer: G has a (normal) subgroup ⟨a⟩ of order p; in fact G ≅ D_p.

Example 7 (UPSC 2024, 10 marks): Let G be a finite group of order mn, where m and n are prime numbers with m > n. Show that G has at most one subgroup of order m.

Suppose H ≠ K are subgroups of order m. H ∩ K is a subgroup of H, so its order divides the prime m, and it is not m (else H = H ∩ K = K). So H ∩ K = {e}.

Then |HK| = o(H)o(K)/o(H ∩ K) = m².

But HK ⊆ G, so m² ≤ mn, i.e. m ≤ n, contradicting m > n.

Answer: G has at most one subgroup of order m (exactly one, by Cauchy's theorem), and it is therefore normal.

Example 8 (UPSC 2025, 10 marks): Let G = {e, x, x², y, yx, yx²} be a non-Abelian group with o(x) = 3 and o(y) = 2. Show that xy = yx² (where e is the identity element of G and o(x), o(y) denote the order of the elements x, y respectively).

xy ∈ G. If xy ∈ {e, x, x²} = ⟨x⟩, then y = x⁻¹(xy) ∈ ⟨x⟩, which is false; so xy ∈ {y, yx, yx²}.

xy = y gives x = e, impossible since o(x) = 3.

xy = yx would make x and y commute; since every element of G is a product of powers of x and y, G would be abelian, contrary to hypothesis.

Hence xy = yx² = yx⁻¹, the defining relation of D₃ ≅ S₃ (one model: x = (1 2 3), y = (1 2), where both sides equal (1 3)).

Answer: xy = yx².

Example 9 (UPSC 2026, 10 marks): Let G be a group of order pq, where p and q are primes. Show that either G is an abelian group or no non-identity element commutes with every element of G.

Z(G) is a subgroup, so by Lagrange o(Z(G)) ∈ {1, p, q, pq}.

If o(Z(G)) = pq, then G = Z(G) is abelian.

If o(Z(G)) = p or q, then o(G/Z(G)) is q or p, a prime, so G/Z(G) is cyclic. Then G is abelian, so Z(G) = G, a contradiction.

Hence either G is abelian, or Z(G) = {e}, which says exactly that no non-identity element commutes with every element of G.

Answer: G is abelian or has trivial centre.

Example 10 (UPSC 2026, 15 marks): Let G be a finite group generated by two elements u and v of order 2 each. Show that G is isomorphic to the dihedral group of order 2n, where O(uv) = n.

Put r = uv and s = u. Then o(r) = n and s² = e; since u⁻¹ = u and v⁻¹ = v, srs⁻¹ = u(uv)u = vu = (uv)⁻¹ = r⁻¹.

v = u(uv) = sr, so G = ⟨u, v⟩ = ⟨r, s⟩. Using sr = r⁻¹s, every word in r, s reduces to rⁱ or rⁱs with 0 ≤ i ≤ n − 1, so G = ⟨r⟩ ∪ ⟨r⟩s and o(G) ≤ 2n.

s ∉ ⟨r⟩: otherwise u and v = sr both lie in the cyclic group ⟨r⟩, which has at most one element of order 2; then u = v, r = e, n = 1, and ⟨r⟩ = {e} cannot contain u. So ⟨r⟩ and ⟨r⟩s are disjoint and o(G) = 2n.

D_n = ⟨ρ, σ⟩ has the same description: 2n elements ρⁱσ^j with o(ρ) = n, σ² = e, σρσ⁻¹ = ρ⁻¹. So ψ(rⁱs^j) = ρⁱσ^j (0 ≤ i ≤ n − 1, j = 0, 1) is a bijection, and it preserves products because in both groups products are computed by the same rules rⁿ = s² = e, srⁱ = r⁻ⁱs.

Answer: G ≅ D_n, the dihedral group of order 2n, with uv as the rotation and u as a reflection.

15. Practice Exercises

Exercise 1 (basic): Find the order of (2, 3) in ℤ₄ × ℤ₉.

Answer: o(2) = 2 in ℤ₄ and o(3) = 3 in ℤ₉, so the order is lcm(2, 3) = 6.

Exercise 2 (basic): Is ℤ₃ × ℤ₁₀ ≅ ℤ₃₀? Is ℤ₆ × ℤ₁₀ ≅ ℤ₆₀?

Answer: yes, since gcd(3, 10) = 1; no, since gcd(6, 10) = 2 and every element of ℤ₆ × ℤ₁₀ has order dividing lcm(6, 10) = 30.

Exercise 3 (basic): Write the class equation of D₅.

Answer: the classes are {e}, {r, r⁴}, {r², r³} and the five reflections, so 10 = 1 + 2 + 2 + 5.

Exercise 4 (standard): List, up to isomorphism, all abelian groups of order 36.

Answer: 36 = 2²·3² gives 2·2 = 4 groups: ℤ₄ × ℤ₉, ℤ₂ × ℤ₂ × ℤ₉, ℤ₄ × ℤ₃ × ℤ₃, ℤ₂ × ℤ₂ × ℤ₃ × ℤ₃.

Exercise 5 (standard): Find the centre of D₆, the symmetry group of a regular hexagon.

Answer: n = 6 is even, so Z(D₆) = {e, r³}, where r³ is the half-turn.

Exercise 6 (standard): Prove that every group of order 33 is cyclic.

Answer: n₁₁ | 3 and n₁₁ ≡ 1 (mod 11) give n₁₁ = 1; n₃ | 11 and n₃ ≡ 1 (mod 3) give n₃ = 1 (11 ≡ 2). The two normal subgroups meet trivially, so G ≅ ℤ₃ × ℤ₁₁ ≅ ℤ₃₃.

Exercise 7 (standard): A simple group G has order 168 = 2³·3·7. How many elements of order 7 does it have?

Answer: n₇ | 24 and n₇ ≡ 1 (mod 7) give n₇ ∈ {1, 8}; simplicity rules out 1. The 8 subgroups of order 7 meet only in e, giving 8·6 = 48 elements of order 7.

Exercise 8 (UPSC level): Prove that no group of order 30 is simple.

Answer: n₅ ∈ {1, 6} and n₃ ∈ {1, 10}. If both were greater than 1 there would be 6·4 = 24 elements of order 5 and 10·2 = 20 of order 3, i.e. 44 > 30 elements. So n₅ = 1 or n₃ = 1, giving a proper non-trivial normal subgroup.

Exercise 9 (UPSC level): Let o(G) = pⁿ with p prime and n ≥ 1. Prove that G has a normal subgroup of order p.

Answer: by the class equation Z(G) ≠ {e}, and p divides o(Z(G)). By Cauchy's theorem Z(G) contains an element z of order p. ⟨z⟩ consists of central elements, so g⟨z⟩g⁻¹ = ⟨z⟩ for all g: it is normal.

Exercise 10 (UPSC level): Show that D₄ and Q₈ are non-isomorphic non-abelian groups of order 8, and that neither is isomorphic to ℤ₂ × ℤ₄.

Answer: count elements of order 2: D₄ has 5, Q₈ has 1, so they are not isomorphic. ℤ₂ × ℤ₄ is abelian (it has 3 elements of order 2), while D₄ and Q₈ are not.

Exercise 11 (UPSC level): Show that a group of order 63 has a normal subgroup of order 7.

Answer: 63 = 3²·7, so n₇ divides 9 and n₇ ≡ 1 (mod 7). Of 1, 3, 9 (≡ 1, 3, 2 mod 7) only 1 works, so the Sylow 7-subgroup is unique and normal.

1. Understanding Rings

Why we need it: integers, polynomials, square matrices and real functions can all be added, subtracted and multiplied, and the two operations are linked by the distributive law. A ring captures exactly this, so facts about factorisation, division and quotients can be proved once for all of them.

In words: '(R, +) is an abelian group' gives a zero 0, negatives −a and the freedom to add in any order. 'Multiplication is associative' lets us write abc without brackets.

The two distributive laws connect the operations, on the left and on the right. Nothing is said about ab = ba, about a unity 1, or about multiplicative inverses; each of these is an extra property.

A unit is an element with a multiplicative inverse. A zero divisor is a non-zero a with ab = 0 or ba = 0 for some non-zero b.

First example, slowly: check that ℤ₆ is a ring and sort its elements.

Step 1: addition mod 6 makes ℤ₆ an abelian group. Step 2: multiplication mod 6 is associative and distributive, because these laws hold in ℤ and reduction mod 6 respects + and ×.

Step 3: 1 is a unity and multiplication commutes. Step 4 (units): 1·1 = 1 and 5·5 = 25 ≡ 1, so the units are 1 and 5.

Step 5 (zero divisors): 2·3 = 6 ≡ 0 and 4·3 = 12 ≡ 0, so 2, 3 and 4 are zero divisors. Notice that each non-zero element is either a unit or a zero divisor.

Non-example: the invertible 2 × 2 real matrices, with matrix addition and multiplication, do not form a ring: I + (−I) = 0 is not invertible, so the set is not closed under addition and has no zero element.

Building up: (1) a·0 = 0 and (−a)(−b) = ab follow from the axioms (Derivation 'Rules of Signs' below). (2) A unit is never a zero divisor: if ab = 0 with a a unit, then b = a⁻¹ab = 0.

(3) In ℤ_n a non-zero k is a unit iff gcd(k, n) = 1 and a zero divisor otherwise. (4) In a ring with unity the characteristic is the additive order of 1.

Where it is used: integral domains and fields are rings with extra properties; polynomial rings, matrix rings and function rings are the main examples of the next topics.

2. Understanding Ideals and Quotient Rings

Why we need it: ℤ_n is built from ℤ by declaring the multiples of n to be zero. To do the same in any ring R we must know which subsets I can be 'set to zero' so that R/I is again a ring. The answer is: the ideals.

In words: an ideal I is an additive subgroup of R that absorbs multiplication by every element of R, not only by elements of I: r·a ∈ I and a·r ∈ I for all r ∈ R and a ∈ I.

The elements of R/I are cosets a + I, added and multiplied through representatives: (a + I)(b + I) = ab + I.

First example, slowly: in R = ℤ × ℤ let I = {(a, 0) : a ∈ ℤ}.

Step 1 (subgroup): (a, 0) − (b, 0) = (a − b, 0) ∈ I, and (0, 0) ∈ I. Step 2 (absorption): for any (r, s) ∈ R, (r, s)(a, 0) = (ra, 0) ∈ I, and the product on the other side is the same since R is commutative.

Step 3: so I is an ideal. The coset (a, b) + I depends only on b, since changing a moves the element inside the same coset.

Step 4: hence (a, b) + I ↦ b is a ring isomorphism R/I ≅ ℤ. What to notice: setting the first coordinate to zero leaves a copy of the second factor.

Non-example: the diagonal D = {(a, a)} is a subring of ℤ × ℤ (it is closed under − and × and contains (1, 1)), but (1, 0)(2, 2) = (2, 0) ∉ D. It does not absorb, so it is not an ideal and (ℤ × ℤ)/D is not a ring.

Building up: (1) The ideal test: I ≠ ∅, a − b ∈ I, and ra, ar ∈ I. (2) Coset multiplication is well defined exactly when I is an ideal (Derivation 'R/I is a Ring Exactly When I is an Ideal' below).

(3) Kernels of ring homomorphisms are ideals. (4) An ideal containing a unit is all of R. (5) Every ideal of ℤ is nℤ, and every ideal of ℤ_n is ⟨d⟩ for a divisor d of n.

Where it is used: quotient rings, prime and maximal ideals, the construction of fields F[x]/⟨p(x)⟩, and the proof that a commutative ring with unity is a field iff it has no proper non-trivial ideals.

3. Understanding Prime and Maximal Ideals

Why we need it: we often want to know whether R/I is an integral domain or a field without computing R/I in detail. Prime and maximal ideals are exactly the ideals for which the answer is yes.

In words: P is prime if P ≠ R and ab ∈ P forces a ∈ P or b ∈ P; this copies 'p | ab ⇒ p | a or p | b'. M is maximal if M ≠ R and no ideal fits strictly between M and R.

The main theorems need R commutative with unity: then R/P is an integral domain iff P is prime, and R/M is a field iff M is maximal.

First example, slowly: take I = 5ℤ in ℤ.

Step 1 (prime): if ab ∈ 5ℤ then 5 | ab, and since 5 is a prime number, 5 | a or 5 | b (Euclid's lemma). So 5ℤ is prime.

Step 2 (maximal): let 5ℤ ⊊ J and pick a ∈ J with 5 ∤ a. Then gcd(a, 5) = 1, so 1 = ax + 5y for some integers x, y (Bézout).

Step 3: ax ∈ J and 5y ∈ 5ℤ ⊆ J, so 1 ∈ J and J = ℤ. So 5ℤ is maximal. Step 4: consistently, ℤ/5ℤ = ℤ₅ is a field.

Non-example: 6ℤ is not prime, since 2·3 ∈ 6ℤ but 2, 3 ∉ 6ℤ, and ℤ₆ has zero divisors. Also {0} ⊂ ℤ is prime but not maximal, because {0} ⊊ 2ℤ ⊊ ℤ; so prime does not imply maximal.

Building up: (1) In R/P, (a + P)(b + P) = P means ab ∈ P, so 'no zero divisors' is word for word 'P is prime' (Standard Proofs, Theorem 11).

(2) If M is maximal and a ∉ M, the ideal M + Ra is bigger than M, so it is R; this gives an inverse of a + M (Theorem 10). (3) A field is an integral domain, so every maximal ideal is prime.

Where it is used: deciding whether R/I is an integral domain or a field, building fields such as ℤ_p and F[x]/⟨p(x)⟩, and the frequently asked 'is R[x]/S an integral domain?' questions.

4. Understanding Ring Homomorphisms

Why we need it: as for groups, the easiest way to identify a quotient ring R/I is to find an onto map from R that respects both operations and has kernel I.

In words: f : R → R′ is a ring homomorphism if f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b. Its kernel {a : f(a) = 0} is an ideal, and R/ker f ≅ f(R).

First example, slowly: let φ : ℤ[x] → ℤ, φ(f) = f(0), the constant term.

Step 1: (f + g)(0) = f(0) + g(0) and (fg)(0) = f(0)g(0), so φ is a ring homomorphism. Step 2: φ(c) = c for every constant c, so φ is onto.

Step 3: f(0) = 0 means every term of f contains x, so f = x·g for some g ∈ ℤ[x]; hence ker φ = ⟨x⟩.

Step 4: by the first isomorphism theorem ℤ[x]/⟨x⟩ ≅ ℤ. Step 5: ℤ is an integral domain but not a field, so ⟨x⟩ is a prime ideal of ℤ[x] that is not maximal.

Non-example: f : ℤ → ℤ, f(n) = 2n, preserves addition, but f(1·1) = 2 while f(1)f(1) = 4. So it is a group homomorphism of (ℤ, +) but not a ring homomorphism.

Building up: (1) A ring homomorphism is in particular an additive group homomorphism, so ker f is a subgroup and f is one-one iff ker f = {0}. (2) If a ∈ ker f and r ∈ R, then f(ra) = f(r)·0 = 0, so the kernel absorbs: it is an ideal.

(3) The group proof of the first isomorphism theorem, with ideals in place of normal subgroups, gives R/ker f ≅ f(R). (4) The kernel of a homomorphism from a field is {0} or the whole field, so such a map is zero or one-one.

Where it is used: ℝ[x]/⟨x² + 1⟩ ≅ ℂ, C[0, 1]/M ≅ ℝ, ℤ_mn ≅ ℤ_m × ℤ_n, the Frobenius map, and imbeddings in Kronecker's theorem.

5. Rings

Ring: a set R with operations + and · such that (R, +) is an abelian group, · is associative, and the distributive laws a(b + c) = ab + ac and (a + b)c = ac + bc hold.

Idea: a ring is arithmetic without division: you can add, subtract and multiply, as in ℤ, polynomials, n × n matrices or real functions. Two familiar habits may fail: multiplication need not commute (matrices), and a product of non-zero elements may be 0 (2·3 = 0 in ℤ₆).

Types: R is commutative if ab = ba for all a, b; R has unity if there is 1 with 1·a = a·1 = a; a unit is an element with a multiplicative inverse.

Basic rules: a·0 = 0·a = 0; a(−b) = (−a)b = −(ab); (−a)(−b) = ab.

Zero divisor: a ≠ 0 such that ab = 0 or ba = 0 for some b ≠ 0. In ℤ_n, a non-zero k is a zero divisor iff gcd(k, n) > 1, and a unit iff gcd(k, n) = 1.

Idea: zero divisors are exactly what breaks cancellation: in ℤ₆, 2·3 = 2·0 but 3 ≠ 0. A ring without them lets you cancel non-zero factors, which is the defining property of an integral domain.

Boolean ring: a ring with a² = a for all a; every Boolean ring is commutative and satisfies a + a = 0.

Characteristic: the least n ≥ 1 with na = 0 for all a ∈ R, and 0 if no such n exists; for a ring with unity it is the additive order of 1.

Why it works: if 1 has additive order n, then n·a = (n·1)a = 0·a = 0 for every a, and no smaller positive multiple kills 1. For ℤ_m × ℤ_n the characteristic is lcm(m, n), the additive order of (1, 1).

6. Subrings and Ideals

Subring test: a non-empty S ⊆ R is a subring iff a − b ∈ S and ab ∈ S for all a, b ∈ S.

Unity of a subring: a subring need not contain the unity of R and may have a unity of its own: in ℤ₆ the subring {0, 2, 4} has unity 4, and in ℤ × ℤ the subring ℤ × {0} has unity (1, 0) ≠ (1, 1). The subring 2ℤ of ℤ has no unity at all.

Ideal: a subring I such that ra ∈ I and ar ∈ I for all r ∈ R and a ∈ I.

Idea: an ideal is a subring that absorbs multiplication from outside, like the even integers: any integer times an even integer is even. This absorption is what makes R/I a ring, exactly as normality makes G/N a group. A subring need not absorb: ℤ is a subring of ℚ, but (1/2)·1 ∉ ℤ.

Ideal test: a non-empty I ⊆ R is an ideal iff a − b ∈ I and ra, ar ∈ I for all a, b ∈ I and r ∈ R.

Principal ideal: in a commutative ring with unity, ⟨a⟩ = aR = {ar : r ∈ R}. Every ideal of ℤ has the form nℤ.

Results: if an ideal contains a unit then I = R; a commutative ring with unity is a field iff its only ideals are {0} and R.

Why it works: if a unit u ∈ I, then 1 = u⁻¹u ∈ I, and then r = r·1 ∈ I for every r ∈ R. So proper ideals contain no units, and a field, where every non-zero element is a unit, has only the ideals {0} and itself (Standard Proofs, Theorem 9, gives the converse).

Operations: I + J and I ∩ J are ideals; I ∪ J is an ideal iff I ⊆ J or J ⊆ I.

Quotient ring: R/I = {a + I : a ∈ R} with (a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = ab + I; these are well-defined because I is an ideal.

7. Prime and Maximal Ideals

Prime ideal: an ideal P ≠ R such that ab ∈ P ⇒ a ∈ P or b ∈ P (R commutative).

Idea: think of an ideal as 'the set of multiples of something'. Then P is prime when divisibility by P behaves like divisibility by a prime number: p | ab ⇒ p | a or p | b. In ℤ, 6ℤ is not prime since 2·3 ∈ 6ℤ but 2, 3 ∉ 6ℤ.

Maximal ideal: an ideal M ≠ R such that if M ⊆ J ⊆ R for an ideal J, then J = M or J = R.

Idea: M is maximal when there is no room between M and R. In ℤ, 6ℤ ⊊ 3ℤ ⊊ ℤ, so 6ℤ is not maximal, while nothing fits between 3ℤ and ℤ.

Key theorem: let R be a commutative ring with unity. Then R/P is an integral domain iff P is prime, and R/M is a field iff M is maximal. Hence every maximal ideal is prime.

Idea: the quotient translates properties of the ideal into properties of a ring: 'P prime' becomes 'R/P has no zero divisors', and 'M maximal' becomes 'R/M has no non-trivial ideals', i.e. R/M is a field. The proofs are in Standard Proofs, Theorems 10 and 11; 'maximal ⇒ prime' then follows because every field is an integral domain.

Why unity is needed: in the ring 2ℤ (no unity), 4ℤ is maximal, because [2ℤ : 4ℤ] = 2 leaves no room between them, but it is not prime, since 2·2 = 4 ∈ 4ℤ while 2 ∉ 4ℤ.

In ℤ: nℤ is prime iff n = 0 or n is prime; nℤ is maximal iff n is prime.

Prime but not maximal: ⟨x⟩ in ℤ[x] is prime since ℤ[x]/⟨x⟩ ≅ ℤ is an integral domain, but it is not maximal since ⟨x⟩ ⊊ ⟨2, x⟩ ⊊ ℤ[x].

8. Ring Homomorphisms

Definition: f : R → R′ with f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b ∈ R.

Idea: a ring homomorphism respects both operations, so it is in particular a homomorphism of the additive groups; everything about kernels and cosets carries over, with ideals playing the role of normal subgroups. It need not send 1 to 1: x ↦ (x, 0) from ℤ to ℤ × ℤ sends 1 to (1, 0).

Kernel: ker f = {a : f(a) = 0} is an ideal of R, and f is one-one iff ker f = {0}.

Isomorphism theorems: (1) R/ker f ≅ f(R); (2) for a subring S and an ideal I, (S + I)/I ≅ S/(S ∩ I); (3) for ideals I ⊆ J, (R/I)/(J/I) ≅ R/J.

Fact: a ring homomorphism from a field is either zero or one-one, because its kernel is an ideal of a field.

Unity under homomorphisms: if φ : R → R′ is onto and R has unity 1, then φ(1) is the unity of R′. If φ maps into an integral domain and φ ≠ 0, then φ(1) = 1′, because φ(1)² = φ(1) ≠ 0 gives φ(1)(φ(1) − 1′) = 0.

Evaluation homomorphisms: for a commutative ring R with unity and c ∈ R, f(x) ↦ f(c) is an onto ring homomorphism R[x] → R with kernel ⟨x − c⟩; so ℤ[x]/⟨x⟩ ≅ ℤ and F[x]/⟨x − c⟩ ≅ F. For c₁ ≠ c₂ in a field F, {f : f(c₁) = f(c₂) = 0} = ⟨(x − c₁)(x − c₂)⟩, and the quotient is ≅ F × F, not an integral domain.

Ring homomorphisms of ℝ and ℤ: the only ring homomorphisms ℝ → ℝ are 0 and the identity, and the same holds for ℤ → ℤ.

Example 1: Show that ℝ[x]/⟨x² + 1⟩ ≅ ℂ.

Define f : ℝ[x] → ℂ by f(p(x)) = p(i). It is a ring homomorphism and onto, since f(a + bx) = a + bi. If p(i) = 0 for a real polynomial p, then p(−i) = 0 too, so x² + 1 divides p; hence ker f = ⟨x² + 1⟩. By the first theorem ℝ[x]/⟨x² + 1⟩ ≅ ℂ, and since ℂ is a field the ideal is maximal.

Example 2: Find all ideals and the maximal ideals of ℤ₁₂.

The ideals are ⟨d⟩ for d dividing 12: ⟨1⟩ = ℤ₁₂, ⟨2⟩, ⟨3⟩, ⟨4⟩, ⟨6⟩, ⟨0⟩. The maximal ideals are ⟨2⟩ and ⟨3⟩, since ℤ₁₂/⟨2⟩ ≅ ℤ₂ and ℤ₁₂/⟨3⟩ ≅ ℤ₃ are fields.

9. Derivation: Rules of Signs from the Axioms

a·0 = 0: a·0 = a(0 + 0) = a·0 + a·0; adding −(a·0) to both sides in the group (R, +) gives a·0 = 0. Similarly 0·a = 0.

a(−b) = −(ab): ab + a(−b) = a(b + (−b)) = a·0 = 0, so a(−b) is the additive inverse of ab. Applying this twice, (−a)(−b) = −(a(−b)) = −(−(ab)) = ab.

Moral: 'minus times minus is plus' is not an extra rule; the distributive law forces it.

10. Derivation: R/I is a Ring Exactly When I is an Ideal

Let I be an additive subgroup, a′ = a + i and b′ = b + j with i, j ∈ I. Then a′b′ = ab + (aj + ib + ij). If I is an ideal, the bracket lies in I, so a′b′ + I = ab + I and the product of cosets is well defined.

Conversely, if the product is well defined, compare (a + I)(0 + I) = I with (a + I)(j + I) = aj + I for j ∈ I: they must agree, so aj ∈ I. In the same way ja ∈ I, so I absorbs multiplication on both sides.

11. Derivation: Ideals of ℤ and ℤ_n, Units and Zero Divisors

Every ideal of ℤ is nℤ: if I ≠ {0}, let n be the least positive element of I. For a ∈ I write a = qn + r with 0 ≤ r < n; then r = a − qn ∈ I, so r = 0 by minimality, and I = nℤ. The same argument with degrees shows every ideal of F[x] is principal.

Ideals of ℤ_n: by the correspondence theorem they are dℤ/nℤ for dℤ ⊇ nℤ, i.e. ⟨d⟩ for the divisors d of n; and ⟨k⟩ = ⟨gcd(k, n)⟩, so ⟨8⟩ = ⟨4⟩ in ℤ₁₂.

Units and zero divisors of ℤ_n: if gcd(k, n) = 1, Bézout gives kx + ny = 1, so kx ≡ 1 and k is a unit. If d = gcd(k, n) > 1 and k ≢ 0, then k·(n/d) = (k/d)·n ≡ 0 with n/d ≢ 0, so k is a zero divisor. Thus every non-zero element of ℤ_n is either a unit or a zero divisor.

12. Choosing the Method

If you must show I is an ideal → use the ideal test (a − b ∈ I, ra and ar ∈ I), or exhibit I as the kernel of a ring homomorphism.

If you must decide whether I is prime or maximal → identify R/I and ask whether it is an integral domain or a field; in ℤ or F[x] look at the generator (prime number, irreducible polynomial).

If all ideals of ℤ_n (or of ℤ_m × ℤ_n) are asked → list ⟨d⟩ for d | n; in a product of rings with unity every ideal is I × J.

If you must identify R/I → define an evaluation or reduction map with kernel I and apply the first isomorphism theorem.

If you must show a ring is not an integral domain or not a field → exhibit a zero divisor, a non-zero non-unit, or a proper non-trivial ideal.

13. Common Mistakes

Calling a subring an ideal: ℤ ⊂ ℚ and the diagonal {(a, a)} ⊂ ℤ × ℤ are subrings but not ideals; always test absorption.

Using 'R/M field ⇔ M maximal' without its hypotheses: R must be commutative with unity; 4ℤ in 2ℤ is maximal but not prime.

Assuming f(1) = 1: a ring homomorphism need not preserve unity unless it is onto or maps into an integral domain and is non-zero.

Saying {0} is never prime: {0} is a prime ideal exactly when R is an integral domain (so {0} is prime in ℤ but not in ℤ₆).

Listing ⟨k⟩ for every k in ℤ_n as different ideals: ⟨k⟩ = ⟨gcd(k, n)⟩, so ℤ_n has exactly as many ideals as n has divisors.

14. Connections

Link: ideals are to rings what normal subgroups are to groups: kernels, quotients and the three isomorphism theorems have the same proofs as in the Normal Subgroups topic.

Link: M_n(ℝ) of Linear Algebra is the standard non-commutative ring, and its zero divisors are the non-zero singular matrices; ℝ[x]/⟨x² + 1⟩ ≅ ℂ (Example 1) is the algebraic construction of the complex numbers used in Complex Analysis.

15. Worked Examples

Example 3: Prove that every Boolean ring R is commutative, and that 2a = 0 for every a ∈ R.

For a ∈ R: a + a = (a + a)² = a² + a² + a² + a² = a + a + a + a, so a + a = 0, i.e. −a = a.

For a, b ∈ R: a + b = (a + b)² = a² + ab + ba + b² = a + ab + ba + b, so ab + ba = 0.

Hence ab = −ba = ba, so R is commutative; if R ≠ {0} its characteristic is 2.

Example 4: Show that the set N of nilpotent elements of a commutative ring R is an ideal of R.

0 ∈ N. Let a^m = 0 and bⁿ = 0. As R is commutative the binomial theorem holds: (a − b)^m+n = ∑_i C(m + n, i)aⁱ(−b)^m+n−i.

In each term either i ≥ m, so aⁱ = 0, or i ≤ m − 1, so m + n − i ≥ n + 1 and b^m+n−i = 0. Hence (a − b)^m+n = 0 and a − b ∈ N.

For r ∈ R, (ra)^m = r^ma^m = 0, so ra ∈ N. Hence N is an ideal (the nilradical of R).

Example 5: Find the characteristic of the ring ℤ₃ × ℤ₄ and decide whether it is an integral domain.

The unity is (1, 1), and k(1, 1) = (k mod 3, k mod 4) = (0, 0) iff 3 | k and 4 | k, i.e. iff 12 | k. So the characteristic is lcm(3, 4) = 12.

It is not an integral domain: (1, 0)·(0, 1) = (0, 0) with both factors non-zero.

This is consistent with the theorem that an integral domain has characteristic 0 or a prime; in fact ℤ₃ × ℤ₄ ≅ ℤ₁₂ as rings.

Example 6: Show that ⟨x² + 1⟩ is not a prime ideal of ℤ₂[x], although it is a maximal ideal of ℤ₃[x].

In ℤ₂[x], (x + 1)² = x² + 2x + 1 = x² + 1, so (x + 1)(x + 1) ∈ I = ⟨x² + 1⟩.

But x + 1 ∉ I, since every non-zero element of I has degree at least 2. So I is not prime, and (x + 1) + I is a zero divisor of ℤ₂[x]/I.

Over ℤ₃, x² + 1 takes the values 1, 2, 2 at 0, 1, 2, so it has no root and is irreducible; ℤ₃[x] is a PID, so the ideal is maximal there.

16. Solved Past UPSC Questions

Example 7 (UPSC 2015, 10 marks): Give an example of a ring having identity but a subring of this having a different identity.

Take R = ℤ₆, whose identity is 1, and S = {0, 2, 4}. For a, b ∈ S, a − b and ab are even residues mod 6, so S is a subring.

In S, 4·0 = 0, 4·2 = 8 = 2 and 4·4 = 16 = 4 (mod 6), so 4 is the identity of S, different from the identity 1 of R.

Another example: R = ℤ × ℤ with identity (1, 1) and the subring S = ℤ × {0}, whose identity is (1, 0).

Answer: ℤ₆ (identity 1) has the subring {0, 2, 4} with identity 4.

Example 8 (UPSC 2015, 15 marks): If R is a ring with unit element 1 and φ is a homomorphism of R onto R′, prove that φ(1) is the unit element of R′.

Let r′ ∈ R′. Since φ is onto, r′ = φ(r) for some r ∈ R.

Then φ(1)r′ = φ(1)φ(r) = φ(1·r) = φ(r) = r′, and similarly r′φ(1) = φ(r·1) = r′.

So φ(1) is a two-sided multiplicative identity of R′, and by uniqueness it is the unit element of R′.

Ontoness is needed: n ↦ (n, 0) is a homomorphism ℤ → ℤ × ℤ with φ(1) = (1, 0), which is not the identity (1, 1).

Answer: φ(1) is the unit element of R′.

Example 9 (UPSC 2018, 20 marks): Suppose ℝ be the set of all real numbers and f : ℝ → ℝ is a function such that the following equations hold for all x, y ∈ ℝ :
(i) f(x + y) = f(x) + f(y)
(ii) f(xy) = f(x) f(y)
Show that ∀ x ∈ ℝ either f(x) = 0, or, f(x) = x.

Put x = y = 1 in (ii): f(1) = f(1)², so f(1) = 0 or f(1) = 1.

If f(1) = 0, then f(x) = f(x·1) = f(x)f(1) = 0 for all x.

Let f(1) = 1. From (i), f(0) = 0, f(−x) = −f(x) and f(nx) = nf(x) for n ∈ ℕ; so f(n) = n for n ∈ ℤ, and nf(m/n) = f(m) = m gives f(q) = q for every q ∈ ℚ.

f preserves order: if t ≥ 0 then t = s² with s = √t, so f(t) = f(s)² ≥ 0. Hence x ≤ y ⇒ f(y) − f(x) = f(y − x) ≥ 0.

Fix x ∈ ℝ. For rationals r < x < s, r = f(r) ≤ f(x) ≤ f(s) = s. Letting r and s tend to x (ℚ is dense) gives f(x) = x.

Answer: either f(x) = 0 for all x, or f(x) = x for all x ∈ ℝ.

Example 10 (UPSC 2022, 15 marks): Let R be a field of real numbers and S, the field of all those polynomials f(x) ∈ R[x] such that f(0) = 0 = f(1). Prove that S is an ideal of R[x]. Is the residue class ring R[x]/S an integral domain? Give justification for your answer.

(The paper says 'the field of all those polynomials'; S is only a subset of ℝ[x], so read 'the set'. Here R = ℝ.)

0 ∈ S. If f, g ∈ S, then (f − g)(0) = 0 and (f − g)(1) = 0, so f − g ∈ S.

If f ∈ S and h ∈ ℝ[x], then (hf)(0) = h(0)f(0) = 0 and (hf)(1) = h(1)f(1) = 0, so hf ∈ S. Hence S is an ideal; in fact S = ⟨x(x − 1)⟩, since x and x − 1 both divide f.

x ∉ S (it is 1 at x = 1) and x − 1 ∉ S (it is −1 at x = 0), but x(x − 1) ∈ S.

So (x + S)(x − 1 + S) = S, the zero of ℝ[x]/S, with both factors non-zero: ℝ[x]/S has zero divisors, and S is not a prime ideal.

Answer: S is an ideal of ℝ[x], but ℝ[x]/S is not an integral domain (indeed ℝ[x]/S ≅ ℝ × ℝ via f + S ↦ (f(0), f(1))).

17. Practice Exercises

Exercise 1 (basic): Find all units and all zero divisors of ℤ₁₀.

Answer: units 1, 3, 7, 9 (gcd with 10 equal to 1); zero divisors 2, 4, 5, 6, 8 (for example 2·5 = 0, 4·5 = 0, 6·5 = 0, 8·5 = 0).

Exercise 2 (basic): Is {0, 3, 6, 9} an ideal of ℤ₁₂? Is it a maximal ideal?

Answer: yes, it is ⟨3⟩. It is maximal, since ℤ₁₂/⟨3⟩ ≅ ℤ₃ is a field.

Exercise 3 (basic): Is the set of real upper triangular 2 × 2 matrices an ideal of M₂(ℝ)?

Answer: no. It is a subring, but [[0, 0], [1, 0]]·[[1, 0], [0, 0]] = [[0, 0], [1, 0]] is not upper triangular, so it does not absorb multiplication from the left.

Exercise 4 (standard): Find all maximal ideals and all prime ideals of ℤ₃₆.

Answer: the ideals are ⟨d⟩ for d | 36, with ℤ₃₆/⟨d⟩ ≅ ℤ_d. This is a field (equivalently, for a finite ring, an integral domain) iff d is prime, so the maximal and the prime ideals are ⟨2⟩ and ⟨3⟩.

Exercise 5 (standard): Show that I = {f ∈ ℤ[x] : f(0) is even} is a maximal ideal of ℤ[x] and that I = ⟨2, x⟩.

Answer: f ↦ f(0) mod 2 is an onto ring homomorphism ℤ[x] → ℤ₂ with kernel I, so ℤ[x]/I ≅ ℤ₂, a field. Writing f = f(0) + x·g shows I = ⟨2, x⟩.

Exercise 6 (standard): Find the characteristic of ℤ₄ × ℤ₆ and of ℤ₃[x].

Answer: lcm(4, 6) = 12 (the additive order of (1, 1)) and 3 (the additive order of the constant 1).

Exercise 7 (standard): In a ring R with unity, prove that (−1)a = −a for every a, and deduce that (−1)(−1) = 1.

Answer: a + (−1)a = (1 + (−1))a = 0·a = 0, so (−1)a is the additive inverse of a. Put a = −1: (−1)(−1) = −(−1) = 1.

Exercise 8 (UPSC level): Prove that ⟨x² + 1⟩ is a maximal ideal of ℝ[x], while ⟨x² − 1⟩ is not even a prime ideal.

Answer: x² + 1 has no real root, so it is irreducible over ℝ and ⟨x² + 1⟩ is maximal in the PID ℝ[x] (indeed ℝ[x]/⟨x² + 1⟩ ≅ ℂ). But (x − 1)(x + 1) ∈ ⟨x² − 1⟩ while neither factor is, having degree 1.

Exercise 9 (UPSC level): Find all ring homomorphisms from ℤ₆ to ℤ₆.

Answer: f(x) = ex with e = f(1); additivity holds for every e, and f(1·1) = f(1)² forces e² = e. The idempotents of ℤ₆ are 0, 1, 3, 4, giving four homomorphisms x ↦ 0, x, 3x, 4x.

Exercise 10 (UPSC level): Show that ℤ[x]/⟨x² + 1⟩ ≅ ℤ[i], and decide whether ⟨x² + 1⟩ is prime or maximal in ℤ[x].

Answer: f ↦ f(i) is onto ℤ[i], and its kernel is ⟨x² + 1⟩ (divide by the monic x² + 1; a remainder a + bx vanishes at i only if a = b = 0). ℤ[i] is an integral domain but not a field, so the ideal is prime but not maximal.

1. Understanding Integral Domains and Fields

Why we need it: when we solve equations we cancel common factors and divide. These steps are legal in ℚ and partly legal in ℤ, but not in ℤ₆, where 2·3 = 2·0 although 3 ≠ 0. We need names for the rings in which they are safe.

In words: an integral domain is a commutative ring with 1 ≠ 0 in which ab = 0 forces a = 0 or b = 0. That is exactly the condition that lets us cancel a non-zero factor.

A field is a commutative ring with 1 ≠ 0 in which every non-zero element has a multiplicative inverse, so that we may also divide.

First example, slowly: show that ℤ₇ is a field. Step 1: ℤ₇ is a commutative ring with unity 1 ≠ 0.

Step 2 (inverses): 1·1 = 1, 2·4 = 8 ≡ 1, 3·5 = 15 ≡ 1 and 6·6 = 36 ≡ 1. So 1⁻¹ = 1, 2⁻¹ = 4, 3⁻¹ = 5, 4⁻¹ = 2, 5⁻¹ = 3 and 6⁻¹ = 6.

Step 3: every non-zero element is a unit, so ℤ₇ is a field. Step 4: it is then automatically an integral domain: if ab = 0 and a ≠ 0, multiply by a⁻¹ to get b = 0.

Non-example: ℤ₈ is not an integral domain, since 2·4 = 8 ≡ 0 with 2, 4 ≠ 0. And ℤ is an integral domain but not a field: 2x = 1 has no integer solution.

Building up: (1) Every field is an integral domain (Step 4 above). (2) Every finite integral domain is a field: x ↦ ax is one-one by cancellation, hence onto, so ax = 1 is solvable (Standard Proofs, Theorem 12).

(3) Hence ℤ_n is a field iff n is prime. (4) The characteristic of an integral domain is 0 or a prime (Theorem 13).

Where it is used: a polynomial of degree n over an integral domain has at most n roots; degrees add in D[x]; quotient fields exist only for integral domains; and finite fields are built from ℤ_p.

2. Understanding Irreducible and Prime Elements

Why we need it: to talk about factorisation in rings other than ℤ we need an analogue of 'prime number'. In ℤ two different properties of primes coincide; in general rings they can separate, and that separation is what breaks unique factorisation.

In words: a unit divides 1, and associates differ by a unit factor (like 3 and −3), so they behave the same for divisibility. 'Irreducible' says p cannot be split: p = ab forces a or b to be a unit.

'Prime' says p behaves well in products: p | ab forces p | a or p | b. Both definitions exclude 0 and the units.

First example, slowly: work in ℤ[i] with the norm N(a + bi) = a² + b², which is multiplicative; the units are the elements of norm 1, namely ±1 and ±i.

Step 1: N(1 + i) = 2. If 1 + i = αβ, then N(α)N(β) = 2, so one of the norms is 1 and that factor is a unit. Hence 1 + i is irreducible.

Step 2: 2 = (1 + i)(1 − i), and N(1 ± i) = 2 ≠ 1, so neither factor is a unit. Hence 2, a prime number in ℤ, is not irreducible in ℤ[i]. Notice that being prime depends on the ring.

Non-example: in ℤ[√−5], 2 is irreducible (no element has norm 2) but not prime: 2 divides (1 + √−5)(1 − √−5) = 6, yet (1 ± √−5)/2 ∉ ℤ[√−5], so 2 divides neither factor.

Building up: (1) In any integral domain a prime is irreducible (the 'Why prime ⇒ irreducible' line below). (2) In a PID an irreducible element is prime (Standard Proofs, Theorem 15).

(3) The norm turns factorisation questions into questions about integers: an element of prime norm is irreducible, and a factor of α has norm dividing N(α).

Where it is used: proving that ℤ[√−5] is not a UFD, factorising Gaussian integers, and showing that F[x]/⟨p(x)⟩ is a field when p is irreducible.

3. Understanding Euclidean Domains, PIDs and UFDs

Why we need it: unique factorisation in ℤ rests on a chain of facts: division with remainder gives gcds, gcds give Euclid's lemma, and Euclid's lemma gives uniqueness. The three classes of rings name the links of this chain, so we can see which link fails in a given ring.

In words: a Euclidean domain has a size function d and a division with smaller remainder: a = bq + r with r = 0 or d(r) < d(b). A PID is one in which every ideal consists of the multiples of a single element.

A UFD is one in which every non-zero non-unit is a product of irreducibles, and any two such products agree after reordering and replacing factors by associates.

First example, slowly: divide 7 + 3i by 2 + i in ℤ[i], with d = N.

Step 1 (exact quotient in ℂ): (7 + 3i)/(2 + i) = (7 + 3i)(2 − i)/5 = (17 − i)/5 = 3.4 − 0.2i.

Step 2 (round to the nearest lattice point): q = 3. Step 3 (remainder): r = (7 + 3i) − 3(2 + i) = 1, with N(r) = 1 < N(2 + i) = 5.

Step 4: so 7 + 3i = 3(2 + i) + 1. The next division, of 2 + i by 1, leaves remainder 0, so the last non-zero remainder 1 is a gcd: 7 + 3i and 2 + i are coprime.

What to notice: rounding each coordinate makes N(α/β − q) at most 1/4 + 1/4 = 1/2, which is why the remainder is always smaller.

Non-example: ℤ[x] is a UFD but not a PID, because ⟨2, x⟩ is not principal (the line 'Why ℤ[x] is not a PID' below). And ℤ[√−5] is not even a UFD: 6 = 2·3 = (1 + √−5)(1 − √−5).

Building up: (1) ED ⇒ PID: in a non-zero ideal, an element of least size generates it (Standard Proofs, Theorem 14). (2) PID ⇒ UFD: chains of principal ideals stop, which gives existence, and irreducible = prime gives uniqueness (the outline 'Every PID is a UFD' below).

(3) Gauss: if D is a UFD so is D[x]; this is why ℤ[x] and F[x, y] are UFDs. (4) Each inclusion of the chain Field ⊂ ED ⊂ PID ⊂ UFD ⊂ integral domain is strict.

Where it is used: gcds and Bézout identities in ℤ[i] and F[x], the minimal polynomial of a matrix or of an algebraic element, and the questions 'prove ℤ[i] is Euclidean' and 'show ℤ[√−5] is not a UFD'.

4. Understanding the Quotient Field

Why we need it: in ℤ the equation 2x = 1 has no solution, and we repair this by passing to fractions. The same repair works for every integral domain: it sits inside a smallest field, its quotient field.

In words: a fraction a/b is a pair (a, b) with b ≠ 0. Two pairs name the same fraction when ad = bc; this test uses only multiplication, which D already has.

Fractions are added and multiplied by the school rules a/b + c/d = (ad + bc)/bd and (a/b)(c/d) = ac/bd.

First example, slowly: take D = ℤ[i] and the fraction 1/(1 + i).

Step 1: (1, 1 + i) ~ (1 − i, 2), because 1·2 = 2 = (1 + i)(1 − i). So 1/(1 + i) = (1 − i)/2 = 1/2 − i/2.

Step 2: in the same way (a + bi)/(c + di) = (a + bi)(c − di)/(c² + d²), which has rational real and imaginary parts. So every fraction lies in ℚ(i) = {p + qi : p, q ∈ ℚ}.

Step 3: conversely p/q + (r/s)i = (ps + qri)/(qs) is a fraction of Gaussian integers. Hence the quotient field of ℤ[i] is ℚ(i).

Non-example: try fractions over ℤ₆, which has zero divisors. (2, 1) ~ (0, 3) since 2·3 = 0 = 1·0, and (0, 3) ~ (0, 1), but (2, 1) ~ (0, 1) fails since 2·1 ≠ 1·0. Transitivity breaks, so zero divisors make the construction impossible.

Building up: (1) ~ is an equivalence relation; transitivity uses cancellation (the line 'Why D must be an integral domain' below). (2) The operations are well defined, and bd ≠ 0 because D has no zero divisors.

(3) The field axioms are checked as for ℚ, with (b, a) the inverse of (a, b) when a ≠ 0. (4) a ↦ a/1 embeds D, and any field containing D contains a copy of these fractions.

Where it is used: ℚ from ℤ, rational functions F(x) from F[x], Gauss's lemma on factorisation over ℚ versus ℤ, and the prime subfield ℚ of a field of characteristic 0.

5. Integral Domains and Fields

Integral domain: a commutative ring with unity 1 ≠ 0 and no zero divisors; equivalently, the cancellation law ab = ac, a ≠ 0 ⇒ b = c holds.

Idea: an integral domain is a ring in which you may cancel, like ℤ; a field is one in which you may also divide. The condition 1 ≠ 0 only excludes the zero ring {0}.

Field: a commutative ring with unity 1 ≠ 0 in which every non-zero element is a unit.

Results: every field is an integral domain; every finite integral domain is a field; ℤ_n is a field iff n is prime.

Idea: in a finite integral domain, x ↦ ax (a ≠ 0) is one-one by cancellation, hence onto by counting, so ax = 1 is solvable (Standard Proofs, Theorem 12). Finiteness is essential: in ℤ the map x ↦ 2x is one-one but not onto.

Characteristic: the characteristic of an integral domain is either 0 or a prime p.

Examples: ℤ[i] = {a + bi : a, b ∈ ℤ} is an integral domain but not a field, since 2 has no inverse; ℤ₆ is not an integral domain, since 2·3 = 0.

6. Divisibility in an Integral Domain

Divides and associates: in an integral domain D, a | b if b = ac for some c ∈ D; a and b are associates if a = ub for a unit u.

Idea: divisibility cannot see units: a and ua divide exactly the same elements and generate the same ideal. In ℤ the associates of 3 are ±3; in ℤ[i] the associates of 1 + 2i are ±(1 + 2i) and ±i(1 + 2i) = ±(−2 + i).

Irreducible element: a non-zero non-unit p such that p = ab implies a or b is a unit.

Prime element: a non-zero non-unit p such that p | ab implies p | a or p | b.

Relation: every prime is irreducible; in a UFD (in particular in a PID) every irreducible is prime.

Why prime ⇒ irreducible: let p be prime and p = ab. Then p | ab, say p | a, so a = pc and p = pcb. Cancelling p (integral domain) gives cb = 1, so b is a unit. For the converse in a PID see Standard Proofs, Theorem 15.

Idea: 'irreducible' says the element cannot be factored; 'prime' says it behaves well inside products. In ℤ they coincide, and the gap between them in rings such as ℤ[√−5] is exactly the failure of unique factorization.

Counter-example: in ℤ[√−5], 6 = 2·3 = (1 + √−5)(1 − √−5). Here 2 is irreducible (no element has norm a² + 5b² = 2) but not prime, since 2 divides the product (1 + √−5)(1 − √−5) but neither factor.

7. Euclidean Domains, PIDs and UFDs

Euclidean domain (ED): an integral domain D with a function d : D∖{0} → {0, 1, 2, ...} such that (i) d(a) ≤ d(ab) for all non-zero a, b, and (ii) for any a and b ≠ 0 there exist q, r ∈ D with a = bq + r, where r = 0 or d(r) < d(b).

Idea: a Euclidean domain is a ring with a division algorithm: d measures size, and each division leaves a smaller remainder. So Euclid's algorithm stops, and gcds exist and can be computed.

Standard EDs: ℤ with d(n) = |n|; F[x] for a field F with d = degree; ℤ[i] with d(a + bi) = a² + b².

Principal ideal domain (PID): an integral domain in which every ideal is principal.

Idea: in a PID every ideal is 'the multiples of one element'. Applied to ⟨a, b⟩ this says gcd(a, b) exists and is a combination ax + by (Bézout). The proof that an ED is a PID (Standard Proofs, Theorem 14) takes an element of least size in the ideal, exactly as for ℤ.

Unique factorization domain (UFD): an integral domain in which every non-zero non-unit is a product of irreducibles, uniquely up to order and associates.

Theorems: every ED is a PID; every PID is a UFD; if D is a UFD then D[x] is a UFD (Gauss).

Strict chain: Field ⊂ ED ⊂ PID ⊂ UFD ⊂ Integral domain. ℤ is an ED but not a field; ℤ[(1 + √−19)/2] is a PID but not an ED; ℤ[x] and F[x, y] are UFDs but not PIDs; ℤ[√−5] is an integral domain but not a UFD.

Maximal ideals in a PID: a non-zero ideal ⟨p⟩ is maximal iff p is irreducible. Hence F[x]/⟨p(x)⟩ is a field iff p(x) is irreducible over F.

Why ℤ[x] is not a PID: if ⟨2, x⟩ = ⟨f⟩, then f divides 2 and x, so f = ±1; but 1 ∉ ⟨2, x⟩, since every element of ⟨2, x⟩ has even constant term.

Example 1: Prove that ℤ[i] is a Euclidean domain.

Let d(α) = N(α) = a² + b² for α = a + bi; N is multiplicative, so N(αβ) = N(α)N(β) ≥ N(α) for β ≠ 0. For α, β ≠ 0 write α/β = x + iy with x, y ∈ ℚ and choose integers m, n with |x − m| ≤ 1/2 and |y − n| ≤ 1/2. Put q = m + in and r = α − βq ∈ ℤ[i]. Then N(r) = N(β)·N(α/β − q) ≤ N(β)(1/4 + 1/4) < N(β).

Example 2: Is ℤ₃[x]/⟨x² + 1⟩ a field?

x² + 1 has no root in ℤ₃ (values at 0, 1, 2 are 1, 2, 2), and a quadratic with no root is irreducible. So the quotient is a field with 3² = 9 elements.

8. Quotient Field

Construction: for an integral domain D, take pairs (a, b) with b ≠ 0 and define (a, b) ~ (c, d) iff ad = bc; this is an equivalence relation, and the class of (a, b) is written a/b.

Idea: this is how ℚ is built from ℤ: a fraction is a pair, and 1/2 = 2/4 because 1·4 = 2·2. The test ad = bc avoids division, which is not yet available in D.

Why D must be an integral domain: for transitivity, let ad = bc and cf = de with b, d, f ≠ 0. Then adf = bcf = bde, so d(af − be) = 0, and since d ≠ 0 and D has no zero divisors, af = be.

Operations: a/b + c/d = (ad + bc)/bd and (a/b)·(c/d) = ac/bd; both are well-defined, and bd ≠ 0 because D has no zero divisors.

Theorem: the classes form a field F, and a ↦ a/1 is a one-one ring homomorphism of D into F.

Minimality: F is the smallest field containing D: any field K containing D contains a subfield isomorphic to F.

Examples: the quotient field of ℤ is ℚ; of ℤ[i] is ℚ(i) = {p + qi : p, q ∈ ℚ}; of F[x] is F(x), the field of rational functions; the quotient field of a field is the field itself.

9. Derivation: Every PID is a UFD (Outline)

Step 1 (chains stop): if ⟨a₁⟩ ⊆ ⟨a₂⟩ ⊆ ... in a PID, the union is an ideal, so it equals ⟨a⟩ for some a; a lies in some ⟨a_n⟩, and then the chain is constant from n on.

Step 2 (existence): if a non-zero non-unit a had no factorization into irreducibles, then a = a₁b₁ with a₁ also having none and ⟨a⟩ ⊊ ⟨a₁⟩; repeating gives a strictly increasing chain, contradicting Step 1.

Step 3 (uniqueness): in a PID every irreducible is prime (Standard Proofs, Theorem 15). If p₁···p_r = q₁···q_s, then p₁ divides some q_j, so p₁ and q_j are associates; cancel and use induction on r.

10. Derivation: The Norm Technique in Quadratic Rings

Multiplicativity: in ℤ[√−n] put N(a + b√−n) = a² + nb² = |a + b√−n|². Since |zw|² = |z|²|w|², N(αβ) = N(α)N(β), and N takes non-negative integer values.

Units: if αβ = 1 then N(α)N(β) = 1, so N(α) = 1; conversely N(α) = 1 means αᾱ = 1. Hence the units of ℤ[√−5] are ±1, and those of ℤ[i] are ±1, ±i.

Irreducibility: if α = βγ with neither a unit, then N(β) is a proper divisor of N(α) greater than 1. So an element of prime norm is irreducible, and 2 ∈ ℤ[√−5] is irreducible because no element has norm 2 (a² + 5b² = 2 has no integer solution).

Primeness through quotients: p is prime iff D/⟨p⟩ is an integral domain. For 3 in ℤ[i], ℤ[i]/⟨3⟩ ≅ ℤ₃[x]/⟨x² + 1⟩, a field with 9 elements, so 3 is prime in ℤ[i]; for 5 = (2 + i)(2 − i) the quotient has zero divisors.

11. Choosing the Method

If the ring is ℤ_n or F[x]/⟨p(x)⟩ → it is a field iff n is prime, or p(x) is irreducible over F.

If you must prove a ring is an ED → give d explicitly and prove the division step by rounding the exact quotient to the nearest lattice point (as for ℤ[i]).

If you must prove a ring is not a PID → exhibit a non-principal ideal such as ⟨2, x⟩ ⊂ ℤ[x] or ⟨x, y⟩ ⊂ F[x, y].

If you must prove a ring is not a UFD → give two factorizations into irreducibles whose factors are not associates, checking irreducibility with the norm.

If a polynomial over ℚ must be shown irreducible → Eisenstein (possibly after x ↦ x + 1), the rational root test for degree ≤ 3, or reduction mod a prime; over a finite field, for degree ≤ 3 check for roots, and for degree 4 also rule out a product of two irreducible quadratics.

If a gcd in ℤ[i] or F[x] is asked → run Euclid's algorithm, rounding quotients in ℤ[i].

12. Common Mistakes

Equating irreducible and prime in every integral domain: they agree in a UFD, but 2 ∈ ℤ[√−5] is irreducible and not prime.

'No root, so irreducible' for degree 4 or more: over ℤ₂, x⁴ + x² + 1 has no root but equals (x² + x + 1)²; over ℚ, x⁴ + 4 = (x² + 2x + 2)(x² − 2x + 2).

Thinking ℤ[x] is a PID because ℤ is: R[x] is a PID only when R is a field.

Ignoring units in uniqueness: 6 = 2·3 = (−2)(−3) is one factorization, not two; uniqueness is only up to order and associates.

Misapplying Eisenstein: all three conditions are needed: p divides every coefficient except the leading one, p does not divide the leading coefficient, and p² does not divide the constant term.

13. Connections

Link: the Gaussian integers are the lattice points of the complex plane and N(z) = |z|² (Complex Analysis); Euclid's algorithm works in ℤ[i] because every point of the plane is within distance √2/2 of a lattice point.

Link: F[x] being a PID is what defines the minimal polynomial of a matrix in Linear Algebra: {p ∈ F[x] : p(A) = 0} is an ideal, generated by the minimal polynomial, which therefore divides the characteristic polynomial (Cayley–Hamilton).

14. Worked Examples

Example 3: Show that 3 is irreducible but not prime in ℤ[√−5].

Use the norm N(a + b√−5) = a² + 5b², which is multiplicative; the units are exactly the elements of norm 1, namely ±1.

If 3 = αβ with neither factor a unit, then N(α)N(β) = 9 forces N(α) = 3; but a² + 5b² = 3 has no integer solution. So 3 is irreducible.

3 divides 6 = (1 + √−5)(1 − √−5), but (1 ± √−5)/3 ∉ ℤ[√−5], so 3 divides neither factor. Hence 3 is not prime, and ℤ[√−5] is not a UFD.

Example 4: Find a gcd of 3 + 4i and 1 + 3i in ℤ[i].

Divide: (3 + 4i)/(1 + 3i) = (3 + 4i)(1 − 3i)/10 = (15 − 5i)/10 = 1.5 − 0.5i. Take q = 1: r = (3 + 4i) − (1 + 3i) = 2 + i, with N(r) = 5 < N(1 + 3i) = 10.

Next, (1 + 3i)/(2 + i) = (1 + 3i)(2 − i)/5 = (5 + 5i)/5 = 1 + i exactly, so the remainder is 0.

The last non-zero remainder 2 + i is a gcd, unique up to the units ±1, ±i. Check: 1 + 3i = (2 + i)(1 + i) and 3 + 4i = (2 + i)².

Example 5: Construct a field with 8 elements.

p(x) = x³ + x + 1 over ℤ₂ has p(0) = 1 and p(1) = 1, so it has no root; a cubic with no root has no linear factor and is therefore irreducible.

ℤ₂[x] is a PID, so ⟨p(x)⟩ is a maximal ideal and F = ℤ₂[x]/⟨p(x)⟩ is a field.

Its elements are a + bα + cα² with a, b, c ∈ ℤ₂, where α = x + ⟨p(x)⟩ satisfies α³ = α + 1. Hence o(F) = 2³ = 8.

Example 6: In an integral domain D, show that ⟨a⟩ = ⟨b⟩ iff a and b are associates.

If a = ub with u a unit, then a ∈ ⟨b⟩ and b = u⁻¹a ∈ ⟨a⟩, so ⟨a⟩ = ⟨b⟩.

Conversely, if ⟨a⟩ = ⟨b⟩, then a = bs and b = at for some s, t ∈ D, so a = a(ts). If a = 0 then b = 0, and a, b are trivially associates.

If a ≠ 0, cancellation (valid in an integral domain) gives ts = 1, so s is a unit and a = sb. Hence a and b are associates.

15. Polynomial Rings, Gcds and Gaussian Integers

Polynomial ring: for a commutative ring R with unity, R[x] = {a₀ + a₁x + ... + a_nxⁿ : a_i ∈ R} with the usual addition and multiplication is a commutative ring with unity 1.

Degree: if R is an integral domain, deg(fg) = deg f + deg g for non-zero f, g, since the product of the leading coefficients is non-zero. Hence R[x] is an integral domain, and its units are exactly the units of R.

With zero divisors this fails: in ℤ₄[x], (1 + 2x)² = 1 + 4x + 4x² = 1, so 1 + 2x is a unit of degree 1.

Division algorithm: if F is a field, f, g ∈ F[x] and g ≠ 0, there are unique q, r ∈ F[x] with f = qg + r and r = 0 or deg r < deg g. Over any commutative ring with unity the same holds when the leading coefficient of g is a unit (for example, g monic).

Remainder and factor theorems: f(x) = (x − c)q(x) + f(c), so c is a root iff (x − c) | f. A non-zero polynomial of degree n over a field (or an integral domain) has at most n roots.

Irreducibility tests: over a field, a polynomial of degree 2 or 3 is irreducible iff it has no root in the field. Over ℚ use Gauss's lemma (a primitive integer polynomial that factors over ℚ factors over ℤ with the same degrees), Eisenstein's criterion, the rational root test (a root p/q in lowest terms has p | a₀ and q | a_n), and reduction mod a prime not dividing the leading coefficient.

Structure of F[x]: F[x] is a Euclidean domain, hence a PID and a UFD; its maximal ideals are ⟨p(x)⟩ with p irreducible, and F[x]/⟨p(x)⟩ is a field with qⁿ elements when F has q elements and deg p = n. ℤ[x] is a UFD but not a PID.

Gcd and Bézout in a PID: in a PID D, ⟨a, b⟩ = ⟨d⟩ with d = gcd(a, b), so d = ax + by for some x, y ∈ D. In ℤ, m₁ℤ + ... + m_kℤ = dℤ with d = gcd(m₁, ..., m_k); for example 4ℤ + 6ℤ = 2ℤ.

Prime ideals and quotients of a PID: in a PID every non-zero prime ideal is maximal, and every ideal of a quotient D/I is principal (it is J/I for an ideal J = ⟨a⟩ ⊇ I, so it equals ⟨a + I⟩).

Gaussian integers: in ℤ[i] the norm N(a + bi) = a² + b² is multiplicative, the units are ±1, ±i (norm 1), and an element of prime norm is irreducible. An odd prime p is irreducible in ℤ[i] iff p ≡ 3 (mod 4); 2 = −i(1 + i)² and 5 = (2 + i)(2 − i) are not irreducible.

16. Solved Past UPSC Questions

Example 7 (UPSC 2014, 15 marks): Show that the set {a + bω : ω³ = 1}, where a and b are real numbers, is a field with respect to usual addition and multiplication.

(Here ω is a non-real cube root of unity, ω = (−1 + i√3)/2; with ω = 1 the set would just be ℝ.)

Let F = {a + bω : a, b ∈ ℝ} ⊆ ℂ. Since ω² + ω + 1 = 0, ω² = −1 − ω. F is closed under subtraction: (a + bω) − (c + dω) = (a − c) + (b − d)ω.

(a + bω)(c + dω) = ac + (ad + bc)ω + bdω² = (ac − bd) + (ad + bc − bd)ω ∈ F. The operations are those of ℂ, so they are associative, commutative and distributive, and 0, 1 ∈ F.

Inverses: (a + bω)((a − b) − bω) = a² − ab − b²(ω + ω²) = a² − ab + b², using ω + ω² = −1.

a² − ab + b² = (a − b/2)² + 3b²/4 > 0 unless a = b = 0. So for a + bω ≠ 0, (a + bω)⁻¹ = ((a − b) − bω)/(a² − ab + b²) ∈ F.

Answer: F is a field; in fact F = ℂ, since a + bω = (a − b/2) + i(√3b/2) takes every complex value.

Example 8 (UPSC 2015, 15 marks): Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields:
(i) The set of numbers of the form b√2 with b rational
(ii) The set of even integers
(iii) The set of positive integers (5 + 6 + 4 = 15)

(i) S = {b√2 : b ∈ ℚ} is closed under + and −, but √2·√2 = 2, and 2 = b√2 would need b = √2 ∉ ℚ. So S is not closed under multiplication: it is not a ring, hence not an integral domain.

(ii) 2ℤ is a commutative ring without zero divisors (ab = 0 in ℤ forces a = 0 or b = 0), but it has no unity: 2e = 2 forces e = 1 ∉ 2ℤ.

So under the usual definition (commutative ring with unity and no zero divisors) 2ℤ is not an integral domain; under Herstein's definition, which does not require unity, it is one. Either way it is not a field, as 2 has no inverse in 2ℤ.

(iii) ℤ⁺ contains no additive identity 0 and no additive inverses, so (ℤ⁺, +) is not a group and ℤ⁺ is not a ring.

Answer: (i) no (not closed under multiplication); (ii) not an integral domain in the sense requiring unity (it is one in Herstein's sense), and not a field; (iii) no.

Example 9 (UPSC 2018, 10 marks): Let R be an integral domain with unit element. Show that any unit in R[x] is a unit in R.

Let f ∈ R[x] be a unit: fg = 1 for some g ∈ R[x], so f, g ≠ 0.

R has no zero divisors, so the product of the leading coefficients of f and g is non-zero and deg(fg) = deg f + deg g.

Hence deg f + deg g = deg 1 = 0, so f = a and g = b are constants of R with ab = 1. Thus f = a is a unit of R.

(The hypothesis is needed: in ℤ₄[x], (1 + 2x)² = 1, so 1 + 2x is a unit that is not in ℤ₄.)

Answer: every unit of R[x] is a constant polynomial that is a unit of R.

Example 10 (UPSC 2020, 10 marks): Let R be a principal ideal domain. Show that every ideal of a quotient ring of R is principal ideal and R/P is a principal ideal domain for a prime ideal P of R.

Let I ≠ R be an ideal of R, π : R → R/I the natural map, and A an ideal of R/I. Then J = π⁻¹(A) = {a ∈ R : a + I ∈ A} is an ideal of R containing I, and π(J) = A because π is onto.

R is a PID, so J = ⟨a⟩. Then every element of A is ra + I = (r + I)(a + I), so A = ⟨a + I⟩ is principal.

Now let P be a prime ideal. R/P is commutative with unity 1 + P ≠ P (as P ≠ R), and (a + P)(b + P) = P ⇒ ab ∈ P ⇒ a ∈ P or b ∈ P, so R/P has no zero divisors.

So R/P is an integral domain in which, by the first part, every ideal is principal. (If P ≠ {0}, P is even maximal and R/P is a field.)

Answer: every ideal of R/I is principal, and R/P is a principal ideal domain for every prime ideal P.

Example 11 (UPSC 2021, 10 marks): Let m₁, m₂, ⋯, m_k be positive integers and d > 0 the greatest common divisor of m₁, m₂, ⋯, m_k. Show that there exist integers x₁, x₂, ⋯, x_k such that d = x₁m₁ + x₂m₂ + ⋯ + x_km_k

Let I = {a₁m₁ + ... + a_km_k : a_i ∈ ℤ}. I is closed under subtraction and under multiplication by integers, so it is an ideal of ℤ, and it contains the positive integer m₁.

Let d′ = x₁m₁ + ... + x_km_k be the least positive element of I. For a ∈ I write a = qd′ + r with 0 ≤ r < d′; then r = a − qd′ ∈ I, so r = 0 by minimality, and I = d′ℤ.

Each m_i ∈ I, so d′ | m_i for every i: d′ is a common divisor, so d′ ≤ d.

Conversely d divides each m_i, hence divides d′ = ∑x_im_i, so d ≤ d′. Therefore d = d′.

Answer: d = x₁m₁ + x₂m₂ + ... + x_km_k for suitable integers x_i; equivalently m₁ℤ + ... + m_kℤ = dℤ.

Example 12 (UPSC 2023, 10 marks): Express the ideal 4Z + 6Z as a principal ideal in the integral domain Z.

4ℤ + 6ℤ = {4a + 6b : a, b ∈ ℤ} is an ideal of ℤ, and ℤ is a PID, so it equals nℤ for its least positive element n.

Every 4a + 6b = 2(2a + 3b) is even, so 4ℤ + 6ℤ ⊆ 2ℤ.

2 = 4(−1) + 6(1) ∈ 4ℤ + 6ℤ, so 2ℤ ⊆ 4ℤ + 6ℤ.

Answer: 4ℤ + 6ℤ = 2ℤ = ⟨2⟩, generated by gcd(4, 6) = 2.

Example 13 (UPSC 2025, 15 marks): Show that 3 is an irreducible element in the integral domain Z[i].

Use the norm N(a + bi) = a² + b². N(αβ) = N(α)N(β), and α is a unit iff N(α) = 1, i.e. α ∈ {±1, ±i}.

3 ≠ 0 and N(3) = 9 ≠ 1, so 3 is a non-zero non-unit.

If 3 = αβ, then N(α)N(β) = 9, so N(α) ∈ {1, 3, 9}.

N(α) = 3 would need a² + b² = 3 in integers, impossible since squares are 0, 1, 4, 9, .... So N(α) = 1 or N(β) = 1: α or β is a unit.

Answer: 3 is irreducible in ℤ[i]; since ℤ[i] is a PID, it is also prime, and ℤ[i]/⟨3⟩ is a field with 9 elements.

Example 14 (UPSC 2026, 10 marks): Show that every non-zero prime ideal in a Euclidean Domain is maximal.

Let D be a Euclidean domain; it is a PID, so a non-zero prime ideal is P = ⟨p⟩ with p ≠ 0, and p is not a unit because P ≠ D.

Let J be an ideal with P ⊆ J ⊆ D. Then J = ⟨a⟩ and p ∈ ⟨a⟩, so p = ab for some b ∈ D.

ab ∈ P and P is prime, so a ∈ P or b ∈ P.

If a ∈ P, then J = ⟨a⟩ ⊆ P, so J = P. If b ∈ P, write b = pc; then p = apc, and cancelling p ≠ 0 gives ac = 1, so a is a unit and J = D.

Answer: P is maximal. ('Non-zero' is needed: ⟨0⟩ is prime in any integral domain but maximal only in a field.)

17. Practice Exercises

Exercise 1 (basic): Is ℤ₂ × ℤ₂ an integral domain?

Answer: no: (1, 0)(0, 1) = (0, 0) with both factors non-zero.

Exercise 2 (basic): List the units of ℤ[i] and the associates of 3 + 2i.

Answer: units ±1, ±i; associates ±(3 + 2i) and ±i(3 + 2i) = ±(−2 + 3i).

Exercise 3 (basic): Find the inverse of 5 in the field ℤ₁₃.

Answer: 8, since 5·8 = 40 = 3·13 + 1.

Exercise 4 (standard): Show that 7 is irreducible in ℤ[i].

Answer: if 7 = αβ with non-units, then N(α) = 7 = a² + b²; but the squares 0, 1, 4 give no sum 7. (In general an odd prime p ≡ 3 (mod 4) stays irreducible.)

Exercise 5 (standard): Factorise 5 and 13 into irreducibles in ℤ[i].

Answer: 5 = (2 + i)(2 − i) and 13 = (3 + 2i)(3 − 2i); each factor has prime norm (5 or 13), so it is irreducible.

Exercise 6 (standard): Find a gcd of 11 + 3i and 1 + 8i in ℤ[i] by Euclid's algorithm.

Answer: 11 + 3i = (1 − i)(1 + 8i) + (2 − 4i); 1 + 8i = (−1 + i)(2 − 4i) + (−1 + 2i); 2 − 4i = (−2)(−1 + 2i). So a gcd is −1 + 2i, an associate of 2 + i (norm 5).

Exercise 7 (standard): Show that x² + x + 2 is irreducible over ℤ₃, and find the number of elements of ℤ₃[x]/⟨x² + x + 2⟩.

Answer: the values at 0, 1, 2 are 2, 1, 2, so there is no root and the quadratic is irreducible. The quotient is a field with 3² = 9 elements.

Exercise 8 (UPSC level): In ℤ[√−5] show that 9 = 3·3 = (2 + √−5)(2 − √−5) are two factorisations into irreducibles that are not related by units.

Answer: N(3) = N(2 ± √−5) = 9, and a proper factor would have norm 3 = a² + 5b², impossible; so all four factors are irreducible. The units are ±1 and 2 ± √−5 ≠ ±3, so the factorisations differ: ℤ[√−5] is not a UFD.

Exercise 9 (UPSC level): Prove that ℤ[√2] = {a + b√2 : a, b ∈ ℤ} is a Euclidean domain with d(a + b√2) = |a² − 2b²|.

Answer: d is multiplicative and is ≥ 1 for non-zero elements (√2 is irrational). Write α/β = x + y√2 with x, y ∈ ℚ, round to integers m, n; then d(α/β − q) = |(x − m)² − 2(y − n)²| ≤ max(1/4, 1/2) = 1/2 < 1, so d(r) < d(β).

Exercise 10 (UPSC level): Show that the quotient field of ℤ[√2] is ℚ(√2) = {a + b√2 : a, b ∈ ℚ}.

Answer: 1/(a + b√2) = (a − b√2)/(a² − 2b²), so every fraction of elements of ℤ[√2] lies in ℚ(√2); conversely p/q + (r/s)√2 = (ps + qr√2)/(qs). ℚ(√2) is a field, so it is the quotient field.

1. Understanding Field Extensions and Degree

Why we need it: the equation x² = 2 has no solution in ℚ, so we must work in a bigger field. We want to measure how much bigger the new field is, and the right measure turns field questions into Linear Algebra.

In words: K is an extension of F when F is a subfield of K. Elements of K can be added and multiplied by elements of F ('scalars'), so K is a vector space over F.

The degree [K : F] is the dimension of this vector space: the number of elements in a basis, possibly infinite.

First example, slowly: K = ℚ(√2) = {a + b√2 : a, b ∈ ℚ} over F = ℚ.

Step 1 (a ring): (a + b√2) + (c + d√2) = (a + c) + (b + d)√2 and (a + b√2)(c + d√2) = (ac + 2bd) + (ad + bc)√2 are again of the same form.

Step 2 (a field): 1/(a + b√2) = (a − b√2)/(a² − 2b²), and the denominator is non-zero because √2 is irrational.

Step 3 (spanning): every element is a·1 + b·√2, so {1, √2} spans K over ℚ. Step 4 (independence): if a + b√2 = 0 with b ≠ 0, then √2 = −a/b would be rational; so b = 0 and then a = 0.

Step 5: {1, √2} is a basis, so [ℚ(√2) : ℚ] = 2.

Non-example: ℤ[√2] = {a + b√2 : a, b ∈ ℤ} contains ℤ but is not a field: 1/2 = a + b√2 would need a = 1/2. So it is not an extension field of ℚ (it does not even contain ℚ).

Building up: (1) The tower law [L : F] = [L : K][K : F] holds because products of basis elements form a basis (Derivation 'The Tower Law' below).

(2) Consequently the degree of any intermediate field divides [L : F]. (3) This divisibility is the tool for 'α ∉ F(β)' questions, such as ∛2 ∉ ℚ(√2).

Where it is used: computing [ℚ(√2, √3) : ℚ], proving that the algebraic elements form a field, subfields of finite fields, and the classical impossibility of ruler-and-compass constructions such as doubling the cube.

2. Understanding Algebraic Elements and Minimal Polynomials

Why we need it: to understand F(α) we want the simplest equation that α satisfies over F. That equation, the minimal polynomial, tells us the degree [F(α) : F], a basis, and how to compute inverses in F(α).

In words: α is algebraic over F if some non-zero polynomial with coefficients in F has α as a root. Its minimal polynomial is the monic one (leading coefficient 1) of least degree.

It is irreducible over F and divides every polynomial in F[x] that vanishes at α. A practical test: a monic irreducible polynomial with root α is the minimal polynomial.

First example, slowly: let α = ∛2 over ℚ. Step 1: α³ − 2 = 0, so α is algebraic.

Step 2: x³ − 2 is irreducible over ℚ by Eisenstein with p = 2: 2 divides the coefficients 0, 0, −2 below the top, 2 does not divide the leading 1, and 4 does not divide −2.

Step 3: it is monic and irreducible with root α, so it is the minimal polynomial, and [ℚ(α) : ℚ] = 3 with basis 1, α, α².

Step 4 (computing in the basis): α⁴ = α·α³ = 2α, and α·α² = 2 gives 1/α = α²/2. What to notice: the relation α³ = 2 reduces every expression to a combination of 1, α, α².

Non-example: √2 is a root of x⁴ − 4, but x⁴ − 4 = (x² − 2)(x² + 2) is reducible, so it is not the minimal polynomial; using it would give the wrong degree 4 instead of 2.

Building up: (1) {f : f(α) = 0} is an ideal of the PID F[x], generated by the minimal polynomial p, which is irreducible (Derivation 'The Minimal Polynomial' below).

(2) Evaluation at α gives F[x]/⟨p⟩ ≅ F(α), so [F(α) : F] = deg p. (3) Every element of a finite extension is algebraic, because 1, α, ..., αⁿ are n + 1 vectors in an n-dimensional space.

Where it is used: finding degrees, writing inverses in F(α), the tower-law arguments of UPSC questions, and the construction of finite fields.

3. Understanding Kronecker's Theorem and Splitting Fields

Why we need it: sometimes no known field contains a root of a polynomial, for example x² + x + 1 over ℤ₂. Kronecker's theorem builds a field in which the root exists, using only F[x] and an irreducible factor.

In words: take an irreducible factor p of f and form K = F[x]/⟨p⟩, which is a field because ⟨p⟩ is maximal. The class α = x + ⟨p⟩ satisfies p(α) = 0, because p itself lies in ⟨p⟩.

F sits inside K as the constant classes a + ⟨p⟩. A splitting field is obtained by repeating this until f factors completely into linear factors.

First example, slowly: F = ℤ₂ and p = x² + x + 1. Step 1: p(0) = 1 and p(1) = 1, so p has no root, and a quadratic without roots is irreducible.

Step 2: K = ℤ₂[x]/⟨p⟩ = {0, 1, α, α + 1}, where α² + α + 1 = 0, i.e. α² = α + 1 (since −1 = 1 in characteristic 2).

Step 3 (products): α·α = α + 1, α(α + 1) = α² + α = 1, and (α + 1)² = α² + 1 = α. So α⁻¹ = α + 1, and every non-zero element is invertible.

Step 4: K is a field with 4 elements, GF(4), and α is a root of p in K. In fact p = (x − α)(x − (α + 1)) there, so K is already the splitting field of p.

Non-example: over ℤ₂, x² + 1 = (x + 1)² is reducible. In ℤ₂[x]/⟨x² + 1⟩ the class β = x + 1 + ⟨x² + 1⟩ is non-zero but β² = 0, so this quotient is not a field. The construction needs an irreducible factor.

Building up: (1) F[x] is a PID, so an irreducible p generates a maximal ideal, and F[x]/⟨p⟩ is a field. (2) a ↦ a + ⟨p⟩ is one-one, since its kernel is an ideal of the field F that misses 1.

(3) Each step adjoins one root and has degree at most the current degree, so the splitting field of a degree-n polynomial has degree at most n! over F (the 'Why at most n!' line below).

Where it is used: the construction of ℂ as ℝ[x]/⟨x² + 1⟩, the existence of GF(pⁿ), and the frequently set Kronecker-type proof question.

4. Understanding Finite Fields

Why we need it: finite fields are the arithmetic of digital codes and of 'clock arithmetic' with a prime modulus. Unlike most algebraic objects, they are completely classified, and their structure answers whole questions at once.

In words: a finite field has prime characteristic p and is a vector space over ℤ_p of some dimension n, hence has pⁿ elements. For each such q = pⁿ there is exactly one field GF(q) up to isomorphism.

Its non-zero elements form a cyclic group of order q − 1, and its subfields are the GF(p^m) with m dividing n.

First example, slowly: GF(9) = ℤ₃[x]/⟨x² + 1⟩ with α² = −1 = 2. Its 9 elements are a + bα with a, b ∈ ℤ₃. We look for a generator of the multiplicative group, which has order 8.

Step 1: α² = 2 and α⁴ = 4 = 1, so o(α) = 4 and α is not a generator.

Step 2: try 1 + α: (1 + α)² = 1 + 2α + α² = 1 + 2α + 2 = 2α, since 3 = 0.

Step 3: (1 + α)⁴ = (2α)² = 4α² = 4·2 = 8 = 2 ≠ 1, so (1 + α)⁸ = 2² = 1.

Step 4: the order divides 8 but not 4, so it is 8, and 1 + α generates GF(9)^*. Notice the shortcut: since 2 is the only prime dividing 8, it was enough to check that the 4th power is not 1.

Non-example: ℤ₉ also has 9 elements but is not a field, since 3·3 = 0. 'The field with 9 elements' is GF(9), not ℤ₉.

Building up: (1) The prime subfield of a finite field is ℤ_p, and a basis of n elements gives pⁿ elements. (2) By Lagrange in K^*, every element satisfies a^q = a, so K is the set of roots of x^q − x.

(3) K^* is cyclic by the φ-counting argument (Derivation 'The Multiplicative Group of a Finite Field is Cyclic' below). (4) The tower law forces subfield degrees to divide n.

Where it is used: generators of GF(q)^* and primitive roots mod p, the Frobenius automorphism, the proof that no finite field is algebraically closed, and error-correcting codes.

5. Prime Subfield and Characteristic

Prime subfield: the smallest subfield of a field K. It is ≅ ℚ if char K = 0 and ≅ ℤ_p if char K = p, a prime.

Why it works: the ring homomorphism ℤ → K, n ↦ n·1, has kernel 0 or pℤ (its image is an integral domain, so the kernel is a prime ideal). In the first case K contains a copy of ℤ and hence of its fraction field ℚ; in the second, a copy of ℤ/pℤ = ℤ_p. Every subfield contains 1, so it contains this copy.

Freshman's dream: in a commutative ring of prime characteristic p, (a + b)^p = a^p + b^p, because p divides the binomial coefficient C(p, k) for 0 < k < p.

Why p | C(p, k): C(p, k) = p!/(k!(p − k)!) is an integer whose numerator contains the prime p, while for 0 < k < p no factor of the denominator does. The rule needs a prime: in ℤ₄ (characteristic 4), (1 + 1)⁴ = 16 = 0 but 1⁴ + 1⁴ = 2.

Frobenius map: in a field K of characteristic p, a ↦ a^p is a one-one ring homomorphism K → K, and an automorphism when K is finite. By Fermat, a^p = a for every a ∈ ℤ_p.

Why it works: it preserves products trivially and sums by the freshman's dream; it is one-one because a^p = 0 ⇒ a = 0 in a field; for finite K one-one implies onto. For infinite K it can fail to be onto: in ℤ_p(t), the element t is not a p-th power.

6. Extensions and Degree

Extension: if F is a subfield of K, then K is an extension of F (written K/F), and K is a vector space over F. Its dimension [K : F] is the degree, and K/F is finite if [K : F] is finite.

Idea: regard K as a vector space over F, so all of Linear Algebra applies and the degree is a dimension. ℂ over ℝ has basis {1, i} and degree 2; ℝ over ℚ has infinite degree, since ℚ is countable while ℝ is not.

Tower law: if F ⊆ K ⊆ L are fields, then [L : F] = [L : K][K : F]; if {α_i} is a basis of K/F and {β_j} a basis of L/K, then {α_iβ_j} is a basis of L/F.

Adjunction: F(α) is the smallest subfield of K containing F and α, and F(α, β) = F(α)(β).

Idea: ℚ(√2) = {a + b√2 : a, b ∈ ℚ}. It is already a field, because 1/(a + b√2) = (a − b√2)/(a² − 2b²) and a² − 2b² ≠ 0 for (a, b) ≠ (0, 0), as √2 is irrational.

Example 1: Find [ℚ(√2, √3) : ℚ] and a basis.

x² − 2 is irreducible over ℚ (Eisenstein, p = 2), so [ℚ(√2) : ℚ] = 2 with basis {1, √2}.

If √3 = a + b√2 with a, b ∈ ℚ, squaring gives 3 = a² + 2b² + 2ab√2, so ab = 0. b = 0 makes √3 rational, and a = 0 makes √(3/2) rational; both are impossible.

So x² − 3 has no root in ℚ(√2) and is irreducible there: [ℚ(√2, √3) : ℚ(√2)] = 2. By the tower law [ℚ(√2, √3) : ℚ] = 4, with basis {1, √2, √3, √6}.

Example 2: Show that ∛2 ∉ ℚ(√2).

x³ − 2 is irreducible over ℚ (Eisenstein, p = 2), so [ℚ(∛2) : ℚ] = 3. If ∛2 ∈ ℚ(√2), then ℚ ⊆ ℚ(∛2) ⊆ ℚ(√2), and by the tower law 3 would divide [ℚ(√2) : ℚ] = 2, which is false.

7. Algebraic Elements

Algebraic and transcendental: α ∈ K is algebraic over F if f(α) = 0 for some non-zero f ∈ F[x]; otherwise α is transcendental (e.g. π and e over ℚ).

Idea: α is algebraic over F when it satisfies a polynomial equation with coefficients in F, like √2, i, ∛2 or e^2πi/n over ℚ. The notion depends on F: π is transcendental over ℚ but algebraic over ℝ, as a root of x − π.

Minimal polynomial: if α is algebraic, the monic polynomial p ∈ F[x] of least degree with p(α) = 0 is unique and irreducible, and it divides every polynomial in F[x] that vanishes at α.

Simple algebraic extension: F(α) ≅ F[x]/⟨p(x)⟩ via x ↦ α, and [F(α) : F] = deg p, with basis 1, α, ..., αⁿ⁻¹; in particular F[α] = F(α).

Why it works: evaluation f ↦ f(α) maps F[x] onto F[α] with kernel ⟨p⟩, and p is irreducible, so ⟨p⟩ is maximal and F[α] ≅ F[x]/⟨p⟩ is a field; hence F[α] = F(α). Every f(α) equals r(α), where r is the remainder of f on division by p and deg r < n, which gives the basis 1, α, ..., αⁿ⁻¹.

Finite ⇒ algebraic: if [K : F] = n, the n + 1 elements 1, α, ..., αⁿ are linearly dependent over F, so every α ∈ K is algebraic, of degree dividing n. Also α is algebraic iff [F(α) : F] is finite.

Closure properties: the elements of K algebraic over F form a subfield of K; and if L/K and K/F are algebraic, then L/F is algebraic.

Example 3: Find the minimal polynomial of α = √2 + √3 over ℚ.

α² = 5 + 2√6, so (α² − 5)² = 24, i.e. α⁴ − 10α² + 1 = 0.

α³ = 11√2 + 9√3, so √2 = (α³ − 9α)/2 ∈ ℚ(α) and √3 = α − √2 ∈ ℚ(α). Hence ℚ(α) = ℚ(√2, √3) has degree 4 over ℚ.

The minimal polynomial has degree 4 and divides x⁴ − 10x² + 1, so it is x⁴ − 10x² + 1.

8. Kronecker's Theorem, Splitting Fields and Algebraic Closure

Kronecker's theorem: every non-constant f ∈ F[x] has a root in some extension of F: take an irreducible factor p of f and K = F[x]/⟨p⟩, in which α = x + ⟨p⟩ is a root.

Idea: if a polynomial has no root, build a field in which it has one: in F[x]/⟨p⟩ the class of x satisfies p by construction. ℂ = ℝ[x]/⟨x² + 1⟩ is the model case, with i = x + ⟨x² + 1⟩.

Splitting field: repeating Kronecker's construction gives an extension of degree at most n! in which f (of degree n) splits into linear factors; the smallest such extension is the splitting field of f, unique up to isomorphism.

Why at most n!: adjoining one root α₁ gives degree ≤ n; over F(α₁), f/(x − α₁) has degree n − 1, so the next step has degree ≤ n − 1, and so on. Example: x³ − 2 over ℚ splits in ℚ(∛2, ω), ω = e^2πi/3, of degree 3·2 = 6 = 3!.

Algebraically closed field: a field in which every non-constant polynomial has a root (equivalently, splits into linear factors). ℂ is algebraically closed (fundamental theorem of algebra); ℝ and ℚ are not, and no finite field is.

Algebraic closure: every field F has an algebraic extension that is algebraically closed; the complex numbers algebraic over ℚ form the algebraic closure of ℚ.

9. Finite Fields

Order: a finite field K has prime characteristic p, is a vector space over ℤ_p of some dimension n, and so has pⁿ elements.

Multiplicative group: the non-zero elements of a finite field form a cyclic group of order pⁿ − 1. In particular U(p) = ℤ_p^* is cyclic.

Existence and uniqueness: for every prime power q = pⁿ there is exactly one field with q elements up to isomorphism, GF(q), the splitting field of x^q − x over ℤ_p; its elements are exactly the roots of x^q − x.

Why the elements are the roots of x^q − x: K^* is a group of order q − 1, so by Lagrange a^q−1 = 1 for a ≠ 0, and a^q = a for every a ∈ K. That gives q distinct roots of a polynomial of degree q, so x^q − x = ∏_a∈K (x − a).

Construction: GF(pⁿ) ≅ ℤ_p[x]/⟨f(x)⟩ for any irreducible f of degree n over ℤ_p; e.g. GF(4) = ℤ₂[x]/⟨x² + x + 1⟩ and GF(9) = ℤ₃[x]/⟨x² + 1⟩.

Subfields: GF(pⁿ) contains a subfield of order p^m iff m divides n, and that subfield is unique.

Why m must divide n: if L is a subfield with p^m elements, the tower law gives n = [GF(pⁿ) : ℤ_p] = [GF(pⁿ) : L]·m. So GF(4) is not a subfield of GF(8), although 4 < 8.

Example 4: Show that every element of GF(8) other than 0 and 1 generates the multiplicative group GF(8)^*.

GF(8)^* has order 8 − 1 = 7, a prime. By Lagrange every element other than 1 has order 7, so it generates the whole group.

10. Derivation: The Tower Law

Spanning: let {α_i} be a basis of K over F and {β_j} a basis of L over K. Any x ∈ L is x = ∑_j c_jβ_j with c_j ∈ K, and each c_j = ∑_i a_ijα_i with a_ij ∈ F. So x = ∑_i,j a_ijα_iβ_j.

Independence: if ∑_i,j a_ijα_iβ_j = 0, group it as ∑_j(∑_i a_ijα_i)β_j = 0. The inner sums lie in K and the β_j are independent over K, so each inner sum is 0; then the α_i are independent over F, so every a_ij = 0.

Consequence: [L : F] = [L : K][K : F]. In particular the degree of any element of K divides [K : F], and if deg α and deg β over F are coprime, [F(α, β) : F] = deg α · deg β.

11. Derivation: The Minimal Polynomial

It exists and divides: I = {f ∈ F[x] : f(α) = 0} is the kernel of evaluation at α, an ideal of F[x], non-zero when α is algebraic. F[x] is a PID, so I = ⟨p⟩; making p monic fixes it uniquely, and every f vanishing at α is a multiple of p.

It is irreducible: if p = gh with deg g, deg h < deg p, then g(α)h(α) = 0 in the field K, so g(α) = 0 or h(α) = 0, contradicting the minimality of deg p.

Practical test: a monic irreducible polynomial with α as a root is the minimal polynomial; so to find deg α, find such a polynomial and prove it irreducible.

12. Derivation: The Multiplicative Group of a Finite Field is Cyclic

Let |K| = q and let ψ(d) be the number of elements of order d in K^*, for d | q − 1. If some a has order d, the d elements of ⟨a⟩ are all roots of x^d − 1, which has at most d roots in a field; so every element of order d lies in ⟨a⟩, and ψ(d) = φ(d). Hence ψ(d) is 0 or φ(d).

Now ∑_d|q−1 ψ(d) = q − 1 = ∑_d|q−1 φ(d) (the count from the Groups topic) and ψ(d) ≤ φ(d) for each d, so ψ(d) = φ(d) for every d. In particular ψ(q − 1) = φ(q − 1) ≥ 1: some element has order q − 1 and generates K^*.

Check: in ℤ₇^*, 3 has powers 3, 2, 6, 4, 5, 1, so 3 is a generator, and there are φ(6) = 2 generators (3 and 5).

13. Choosing the Method

If [F(α) : F] is asked → find a monic polynomial with root α and prove it irreducible (Eisenstein, rational roots, no root for degree ≤ 3).

If [F(α, β) : F] is asked → tower law; show the minimal polynomial of β stays irreducible over F(α), or use coprime degrees.

If you must show α ∉ F(β) → compare degrees: deg α must divide [F(β) : F].

If you must show an element is algebraic → produce a polynomial it satisfies, or show it lies in a finite extension.

If you must construct GF(pⁿ) → find an irreducible polynomial of degree n over ℤ_p (for n = 2, 3 one with no root) and take ℤ_p[x]/⟨f⟩.

If a generator of GF(q)^* is asked → check that a^(q−1)/r ≠ 1 for each prime r dividing q − 1.

14. Common Mistakes

Taking the degree of any polynomial satisfied by α: √2 satisfies x⁴ − 4, but its degree over ℚ is 2; only the irreducible one counts.

Multiplying degrees blindly: [ℚ(√2, √8) : ℚ] = 2, not 4, because √8 = 2√2 ∈ ℚ(√2).

Confusing ℤ_pⁿ with GF(pⁿ): ℤ₄ is not a field (2·2 = 0); GF(4) = ℤ₂[x]/⟨x² + x + 1⟩.

Using (a + b)ⁿ = aⁿ + bⁿ outside its range: it holds for n = p (and its powers) in characteristic p only.

Claiming a minimal polynomial without proving irreducibility: examiners deduct marks; give the reason (Eisenstein, no rational root, degree argument).

15. Connections

Link: degree is dimension (Linear Algebra): the tower law is a statement about bases, and the minimal polynomial of α over F is the minimal polynomial of the linear map x ↦ αx of K.

Link: the fundamental theorem of algebra (ℂ is algebraically closed) is proved in Complex Analysis with Liouville's theorem; finite fields are behind the error-correcting codes and the binary arithmetic of the Computer Programming chapter (GF(2) = {0, 1} with XOR as addition).

16. Solved Past UPSC Questions

Example 5 (UPSC 2016, 20 marks): Let K be an extension of a field F. Prove that the elements of K, which are algebraic over F, form a subfield of K. Further, if F ⊂ K ⊂ L are fields, L is algebraic over K and K is algebraic over F, then prove that L is algebraic over F.

Let α, β ∈ K be algebraic over F. [F(α) : F] is finite; β is algebraic over F, hence over F(α) (the same polynomial works), so [F(α, β) : F(α)] is finite.

By the tower law [F(α, β) : F] = [F(α, β) : F(α)][F(α) : F] is finite.

α ± β, αβ and α/β (β ≠ 0) lie in F(α, β), and every element of a finite extension is algebraic. So these are algebraic over F, and with 0, 1 the algebraic elements form a subfield of K.

Now let L/K and K/F be algebraic and γ ∈ L. Then γⁿ + a_n−1γⁿ⁻¹ + ... + a₀ = 0 for some a_i ∈ K.

Put E = F(a₀, ..., a_n−1). Each a_i is algebraic over F, so adjoining them one at a time and using the tower law shows [E : F] is finite; and γ is algebraic over E by the same equation, so [E(γ) : E] ≤ n.

Hence [E(γ) : F] = [E(γ) : E][E : F] is finite, so γ is algebraic over F.

Answer: the algebraic elements of K form a subfield, and an algebraic extension of an algebraic extension is algebraic (L/F is algebraic).

Example 6 (UPSC 2016, 15 marks): Show that every algebraically closed field is infinite.

Let F be algebraically closed and suppose F is finite, F = {a₁, a₂, ..., a_n}.

Consider f(x) = (x − a₁)(x − a₂)···(x − a_n) + 1 ∈ F[x], of degree n ≥ 2 (a field has 0 ≠ 1).

For every i, f(a_i) = 0 + 1 = 1 ≠ 0, so f has no root in F, contradicting algebraic closure.

Answer: every algebraically closed field is infinite.

Example 7 (UPSC 2020, 15 marks): Let R be a finite field of characteristic p (> 0). Show that the mapping f : R → R defined by f(a) = a^p, ∀a ∈ R is an isomorphism.

f(ab) = (ab)^p = a^pb^p = f(a)f(b), since R is commutative.

(a + b)^p = ∑_k=0^p C(p, k)a^kb^p−k. For 0 < k < p, C(p, k) = p!/(k!(p − k)!) is divisible by p (p divides the numerator but not k!(p − k)!), so these terms vanish in characteristic p, and f(a + b) = a^p + b^p = f(a) + f(b).

One-one: f(a) = 0 ⇒ a^p = 0 ⇒ a = 0, because a field has no zero divisors; so ker f = {0}. Also f(1) = 1.

Onto: R is finite and f : R → R is one-one, so f is onto.

Answer: f(a) = a^p is a ring isomorphism of R onto itself (the Frobenius automorphism).

Example 8 (UPSC 2021, 15 marks): Let F be a field and f(x) ∈ F[x] a polynomial of degree > 0 over F. Show that there is a field F′ and an imbedding q : F → F′ s.t. the polynomial f^q ∈ F′[x] has a root in F′, where f^q is obtained by replacing each coefficient a of f by q(a).

F[x] is a UFD and deg f > 0, so f = p·g with p ∈ F[x] irreducible.

F[x] is a PID and p is irreducible, so ⟨p⟩ is a maximal ideal and F′ = F[x]/⟨p⟩ is a field.

Define q : F → F′ by q(a) = a + ⟨p⟩. It is a ring homomorphism with q(1) = 1 + ⟨p⟩ ≠ 0 (deg p ≥ 1), so its kernel is an ideal of the field F other than F, i.e. {0}: q is an imbedding.

Let α = x + ⟨p⟩ ∈ F′. If p = ∑c_ixⁱ, then p^q(α) = ∑(c_i + ⟨p⟩)(x + ⟨p⟩)ⁱ = p(x) + ⟨p⟩ = 0 in F′.

h ↦ h^q is a ring homomorphism F[x] → F′[x], so f^q = p^qg^q and f^q(α) = p^q(α)g^q(α) = 0.

Answer: F′ = F[x]/⟨p(x)⟩ with the imbedding a ↦ a + ⟨p⟩ works, and α = x + ⟨p(x)⟩ is a root of f^q (Kronecker's theorem).

17. Practice Exercises

Exercise 1 (basic): Find [ℚ(i) : ℚ], a basis, and the minimal polynomial of i over ℚ.

Answer: x² + 1 is monic, irreducible over ℚ (no rational, indeed no real, root) and has root i; so the degree is 2 with basis {1, i}.

Exercise 2 (basic): Is ℚ(√2) = ℚ(√8)? Is ℚ(√2) = ℚ(√3)?

Answer: yes, since √8 = 2√2; no, since √3 ∉ ℚ(√2) (squaring √3 = a + b√2 forces ab = 0, and both cases are impossible).

Exercise 3 (basic): State the characteristic and the prime subfield of ℤ₅(t), the field of rational functions over ℤ₅.

Answer: characteristic 5, prime subfield ℤ₅; the field is infinite although its characteristic is not 0.

Exercise 4 (standard): Find the minimal polynomial of 1 + √3 over ℚ.

Answer: (x − 1)² = 3 gives x² − 2x − 2, irreducible by Eisenstein with p = 2; so it is the minimal polynomial and the degree is 2.

Exercise 5 (standard): Find [ℚ(√2, ∛2) : ℚ].

Answer: 6. The degrees 2 and 3 of √2 and ∛2 both divide the total degree, which is at most 2·3; since gcd(2, 3) = 1 it equals 6.

Exercise 6 (standard): Let α be a real root of x³ − x − 1. Show that this polynomial is irreducible over ℚ and express α⁻¹ in the basis 1, α, α².

Answer: the only possible rational roots ±1 give −1 and −1, so the cubic has no root and is irreducible. From α³ − α = 1, α(α² − 1) = 1, so α⁻¹ = α² − 1.

Exercise 7 (standard): In GF(8) = ℤ₂[x]/⟨x³ + x + 1⟩ with α = x + ⟨x³ + x + 1⟩, list the powers α, α², ..., α⁷.

Answer: α, α², α³ = α + 1, α⁴ = α² + α, α⁵ = α² + α + 1, α⁶ = α² + 1, α⁷ = 1: all seven non-zero elements appear, so α generates GF(8)^*.

Exercise 8 (UPSC level): Prove that GF(16) has exactly one subfield with 4 elements and no subfield with 8 elements.

Answer: a subfield of order 2^m needs m | 4 by the tower law, so 8 = 2³ is impossible. For m = 2, the subfield is {a : a⁴ = a}, the roots of x⁴ − x, so there is exactly one; it exists because GF(16)^* is cyclic of order 15 and has a unique subgroup of order 3.

Exercise 9 (UPSC level): Find the degree over ℚ of the splitting field of x⁴ − 2.

Answer: 8. The roots are ±⁴√2 and ±i⁴√2, so the splitting field is ℚ(⁴√2, i). x⁴ − 2 is Eisenstein at 2, giving degree 4 for ℚ(⁴√2) ⊂ ℝ, and i ∉ ℝ adds a factor 2.

Exercise 10 (UPSC level): Let α be algebraic over F with [F(α) : F] odd. Prove that F(α²) = F(α).

Answer: α is a root of x² − α² over F(α²), so [F(α) : F(α²)] ≤ 2. By the tower law it divides the odd number [F(α) : F], so it is 1.

Exercise 11 (UPSC level): Let F be a field of characteristic p and a ∈ F. Prove that x^p − a has at most one root in any extension of F.

Answer: if b is a root, then by the freshman's dream (x − b)^p = x^p − b^p = x^p − a, so every root c satisfies (c − b)^p = 0, i.e. c = b.

1. Lagrange's Theorem

Statement: Let G be a finite group and H a subgroup of G. Then o(H) divides o(G), and [G : H] = o(G)/o(H).

Proof: Define a ~ b on G by a⁻¹b ∈ H. It is reflexive (a⁻¹a = e ∈ H), symmetric (b⁻¹a = (a⁻¹b)⁻¹ ∈ H) and transitive (a⁻¹c = (a⁻¹b)(b⁻¹c) ∈ H), so it is an equivalence relation.

The class of a is {b : a⁻¹b ∈ H} = {ah : h ∈ H} = aH. Hence the distinct left cosets are pairwise disjoint and their union is G.

For each a, the map h ↦ ah from H to aH is onto by definition and one-one by left cancellation (ah₁ = ah₂ ⇒ h₁ = h₂). So |aH| = |H|.

Since G is finite there are finitely many distinct cosets, say a₁H, ..., a_kH with k = [G : H], and G = a₁H ∪ ... ∪ a_kH is a disjoint union.

Counting elements, o(G) = k·o(H). Therefore o(H) divides o(G) and [G : H] = o(G)/o(H).

Hence proved.

2. Every Subgroup of a Cyclic Group is Cyclic

Statement: Let G = ⟨a⟩ be a cyclic group and H a subgroup of G. Then H is cyclic.

Proof: If H = {e}, then H = ⟨e⟩ is cyclic. Otherwise H contains some a^k with k ≠ 0, and then also a^−k, so H contains a positive power of a.

By the well-ordering principle, let m be the least positive integer with a^m ∈ H. Since H is closed, ⟨a^m⟩ ⊆ H.

Let aⁿ ∈ H. By the division algorithm, n = qm + r with 0 ≤ r ≤ m − 1.

Then a^r = aⁿ(a^m)^−q ∈ H. Since r is less than m, the minimality of m forces r = 0.

So aⁿ = (a^m)^q ∈ ⟨a^m⟩, which gives H ⊆ ⟨a^m⟩. Hence H = ⟨a^m⟩ is cyclic.

Hence proved.

3. A Group of Prime Order is Cyclic

Statement: Let G be a group of order p, where p is a prime. Then G is cyclic, every non-identity element generates G, and G ≅ ℤ_p.

Proof: Since p ≥ 2, G has an element a ≠ e. Let H = ⟨a⟩, a subgroup of G.

By Lagrange's theorem o(H) divides p, so o(H) = 1 or p.

H contains the two distinct elements e and a, so o(H) ≠ 1. Hence o(H) = p = o(G), and since H ⊆ G, H = G.

Thus G = ⟨a⟩ for every a ≠ e, and a cyclic group of order p is isomorphic to ℤ_p (map a^k ↦ k mod p). In particular G is abelian.

Hence proved.

4. A Subgroup of Index 2 is Normal

Statement: Let H be a subgroup of a group G (finite or infinite) with [G : H] = 2. Then H ⊴ G.

Proof: Let g ∈ G. If g ∈ H, then gH = H = Hg.

If g ∉ H, then gH ≠ H. There are only two left cosets and they partition G, so gH = G ∖ H.

Likewise there are only two right cosets, H and Hg ≠ H (as g ∈ Hg but g ∉ H), so Hg = G ∖ H.

Thus gH = Hg for every g ∈ G, which is the definition of a normal subgroup.

Hence proved.

5. The Kernel of a Homomorphism

Statement: Let f : G → G′ be a group homomorphism and K = ker f = {a ∈ G : f(a) = e′}. Then K ⊴ G, and f is one-one iff K = {e}.

Proof: f(e) = f(ee) = f(e)f(e), and cancelling gives f(e) = e′. Then f(a)f(a⁻¹) = f(e) = e′, so f(a⁻¹) = f(a)⁻¹.

So e ∈ K and K is non-empty. For a, b ∈ K, f(ab⁻¹) = f(a)f(b)⁻¹ = e′, so ab⁻¹ ∈ K and K ≤ G by the one-step test.

For g ∈ G and k ∈ K, f(gkg⁻¹) = f(g)e′f(g)⁻¹ = e′, so gKg⁻¹ ⊆ K and K ⊴ G.

If f is one-one and a ∈ K, then f(a) = e′ = f(e) gives a = e; so K = {e}.

Conversely, if K = {e} and f(a) = f(b), then f(ab⁻¹) = e′, so ab⁻¹ = e and a = b; so f is one-one.

Hence proved.

6. Fundamental Theorem of Homomorphism (First Isomorphism Theorem)

Statement: Let f : G → G′ be a group homomorphism with kernel K. Then G/K ≅ f(G). In particular, if f is onto, G/K ≅ G′.

Proof: By the previous theorem K ⊴ G, so G/K is a group under (aK)(bK) = abK.

Define φ : G/K → f(G) by φ(aK) = f(a).

Well-defined and one-one: aK = bK ⇔ a⁻¹b ∈ K ⇔ f(a)⁻¹f(b) = e′ ⇔ f(a) = f(b). The implication ⇒ shows φ is well defined, and ⇐ shows φ is one-one.

Homomorphism: φ(aK·bK) = φ(abK) = f(ab) = f(a)f(b) = φ(aK)φ(bK).

Onto: every element of f(G) is f(a) = φ(aK) for some a ∈ G.

So φ is an isomorphism and G/K ≅ f(G); if f is onto, f(G) = G′.

Hence proved.

7. Cayley's Theorem

Statement: Every group G is isomorphic to a subgroup of Sym(G), the group of all permutations of the set G. In particular, every group of order n is isomorphic to a subgroup of S_n.

Proof: For a ∈ G define T_a : G → G by T_a(x) = ax.

T_a is one-one, since ax = ay ⇒ x = y, and onto, since T_a(a⁻¹y) = y. So T_a ∈ Sym(G).

Define φ : G → Sym(G) by φ(a) = T_a. For all x, (T_a ∘ T_b)(x) = a(bx) = (ab)x = T_ab(x), so φ(ab) = φ(a)φ(b) and φ is a homomorphism.

If φ(a) = T_a is the identity map, then a = ae = T_a(e) = e. So ker φ = {e} and φ is one-one.

By the first isomorphism theorem, G ≅ φ(G) = {T_a : a ∈ G}, which is a subgroup of Sym(G). If o(G) = n, labelling the elements 1, ..., n identifies Sym(G) with S_n.

Hence proved.

8. A_n is a Normal Subgroup of S_n of Index 2

Statement: For n ≥ 2, the set A_n of even permutations is a normal subgroup of S_n, and o(A_n) = n!/2. (Assume the parity theorem: no permutation is both even and odd.)

Proof: Define sgn : S_n → {1, −1} (a group under multiplication) by sgn σ = 1 if σ is even and −1 if σ is odd. By the parity theorem this is well defined.

If σ is a product of r transpositions and τ of s transpositions, then στ is a product of r + s transpositions, so sgn(στ) = (−1)^r+s = sgn σ · sgn τ. Thus sgn is a homomorphism.

sgn is onto, since sgn(e) = 1 and sgn((1 2)) = −1 (this needs n ≥ 2).

ker sgn = {σ : σ even} = A_n, so A_n ⊴ S_n by the kernel theorem.

By the first isomorphism theorem S_n/A_n ≅ {1, −1}, so [S_n : A_n] = 2 and o(A_n) = n!/2.

Hence proved.

9. A Commutative Ring with Unity is a Field iff it has no Proper Non-trivial Ideals

Statement: Let R be a commutative ring with unity 1 ≠ 0. Then R is a field iff the only ideals of R are {0} and R.

Proof: (⇒) Let R be a field and I a non-zero ideal. Pick a ∈ I with a ≠ 0. Then 1 = a⁻¹a ∈ I, so r = r·1 ∈ I for every r ∈ R, and I = R.

(⇐) Suppose the only ideals are {0} and R, and let a ∈ R, a ≠ 0. Consider aR = {ar : r ∈ R}.

aR is an ideal: ar − as = a(r − s) ∈ aR, and t(ar) = a(rt) ∈ aR for t ∈ R, using commutativity.

aR contains a = a·1 ≠ 0, so aR ≠ {0} and hence aR = R.

In particular 1 ∈ aR, so 1 = ab for some b ∈ R. Thus every non-zero element is a unit, and R is a field.

Hence proved.

10. R/M is a Field iff M is a Maximal Ideal

Statement: Let R be a commutative ring with unity and M an ideal of R. Then R/M is a field iff M is a maximal ideal.

Proof: R/M is a commutative ring with unity 1 + M, and 1 + M ≠ 0 + M iff 1 ∉ M iff M ≠ R.

(⇒) Let R/M be a field; then M ≠ R. Let J be an ideal with M ⊊ J, and pick a ∈ J ∖ M. Then a + M ≠ M, so there is b with (a + M)(b + M) = 1 + M, i.e. 1 − ab ∈ M ⊆ J.

Since ab ∈ J, 1 = (1 − ab) + ab ∈ J, so J = R. Hence M is maximal.

(⇐) Let M be maximal, so M ≠ R and 1 + M ≠ 0 + M. Take a + M ≠ M, i.e. a ∉ M, and put J = M + Ra = {m + ra : m ∈ M, r ∈ R}.

J is an ideal ((m + ra) − (m′ + r′a) = (m − m′) + (r − r′)a and s(m + ra) = sm + (sr)a), it contains M, and it contains a = 0 + 1·a ∉ M. By maximality J = R.

So 1 = m + ba for some m ∈ M, b ∈ R, and (b + M)(a + M) = ba + M = (1 − m) + M = 1 + M. Every non-zero element of R/M is a unit, so R/M is a field.

Hence proved.

11. R/P is an Integral Domain iff P is a Prime Ideal

Statement: Let R be a commutative ring with unity and P an ideal of R. Then R/P is an integral domain iff P is a prime ideal. Consequently every maximal ideal is prime.

Proof: R/P is a commutative ring with unity 1 + P, and 1 + P ≠ 0 + P iff P ≠ R.

The zero of R/P is P, and (a + P)(b + P) = ab + P, so (a + P)(b + P) = P ⇔ ab ∈ P, while a + P = P ⇔ a ∈ P.

Hence 'R/P has no zero divisors' reads: ab ∈ P ⇒ a ∈ P or b ∈ P.

So R/P is an integral domain ⇔ P ≠ R and [ab ∈ P ⇒ a ∈ P or b ∈ P] ⇔ P is a prime ideal.

If M is maximal, R/M is a field by the previous theorem; a field is an integral domain, so M is prime.

Hence proved.

12. Every Finite Integral Domain is a Field

Statement: Let D be a finite integral domain. Then D is a field. In particular ℤ_p is a field for every prime p.

Proof: D is commutative with unity 1 ≠ 0, so it suffices to show every a ≠ 0 is a unit.

Fix a ≠ 0 and define λ_a : D → D by λ_a(x) = ax.

λ_a is one-one: ax = ay ⇒ a(x − y) = 0 ⇒ x − y = 0, because a ≠ 0 and D has no zero divisors.

A one-one map of a finite set into itself is onto, so ab = 1 for some b ∈ D. Hence a is a unit and D is a field.

For a prime p, ℤ_p has no zero divisors (p | ab ⇒ p | a or p | b), so it is a finite integral domain and therefore a field.

Hence proved.

13. The Characteristic of an Integral Domain is 0 or a Prime

Statement: Let D be an integral domain. Then the characteristic of D is either 0 or a prime number p.

Proof: For any a ∈ D and n ∈ ℕ, n·a = (n·1)a. So if n·1 = 0 then n·a = 0 for all a, and the characteristic equals the additive order of 1 (or 0 if that order is infinite).

Suppose the characteristic is n > 0, so n is the least positive integer with n·1 = 0. Since 1 ≠ 0, n ≠ 1.

If n were composite, n = rs with 1 < r, s < n. By the distributive law 0 = n·1 = (r·1)(s·1).

As D has no zero divisors, r·1 = 0 or s·1 = 0, contradicting the minimality of n.

Hence n is prime; otherwise the characteristic is 0.

Hence proved.

14. Every Euclidean Domain is a PID

Statement: Let D be a Euclidean domain with Euclidean function d. Then every ideal of D is principal, so D is a PID. In particular ℤ, F[x] (F a field) and ℤ[i] are PIDs.

Proof: Let I be an ideal of D. If I = {0}, then I = ⟨0⟩ is principal.

Otherwise the set {d(x) : x ∈ I, x ≠ 0} is a non-empty set of non-negative integers; by well-ordering choose b ∈ I, b ≠ 0, with d(b) least. Since I is an ideal, ⟨b⟩ ⊆ I.

Let a ∈ I. By the division property, a = bq + r with r = 0 or d(r) < d(b).

r = a − bq ∈ I because a, bq ∈ I. If r ≠ 0, then d(r) < d(b) contradicts the choice of b; so r = 0.

Hence a = bq ∈ ⟨b⟩, so I ⊆ ⟨b⟩ and I = ⟨b⟩. Every ideal is principal, and D is a PID.

Hence proved.

15. In a PID, Irreducible Elements Generate Maximal Ideals and are Prime

Statement: Let D be a PID and p ∈ D a non-zero non-unit. Then ⟨p⟩ is a maximal ideal iff p is irreducible. Consequently every irreducible element of D is prime.

Proof: ⟨p⟩ ≠ D, because p is not a unit (1 ∉ ⟨p⟩).

(⇐) Let p be irreducible and ⟨p⟩ ⊆ J for an ideal J. As D is a PID, J = ⟨a⟩, so p = ab for some b ∈ D.

By irreducibility, either a is a unit, giving J = D, or b is a unit, giving a = pb⁻¹ ∈ ⟨p⟩ and J = ⟨p⟩. So ⟨p⟩ is maximal.

(⇒) Let ⟨p⟩ be maximal and p = ab with a not a unit. Then ⟨p⟩ ⊆ ⟨a⟩ ≠ D, so ⟨a⟩ = ⟨p⟩ and a = pc for some c.

Then p = pcb, and cancelling p ≠ 0 gives cb = 1, so b is a unit. Hence p is irreducible.

Finally, if p is irreducible, ⟨p⟩ is maximal, hence prime. So p | ab ⇒ ab ∈ ⟨p⟩ ⇒ a ∈ ⟨p⟩ or b ∈ ⟨p⟩ ⇒ p | a or p | b, i.e. p is a prime element.

Hence proved.

16. A Finite Cyclic Group of Order n has φ(n) Generators

Statement: Let G = ⟨a⟩ be a cyclic group of order n. Then a^k generates G iff gcd(k, n) = 1, so G has exactly φ(n) generators.

Proof: The elements a¹, a², ..., aⁿ = e are distinct, since aⁱ = a^j with 1 ≤ i < j ≤ n would give a^j−i = e with 0 < j − i < n; so G = {a^k : 1 ≤ k ≤ n}.

Let 1 ≤ k ≤ n and d = gcd(k, n). Then (a^k)^n/d = (aⁿ)^k/d = e.

If (a^k)^m = e, then n | km, so (n/d) | (k/d)m; since gcd(n/d, k/d) = 1, (n/d) | m. Hence o(a^k) = n/d.

⟨a^k⟩ has o(a^k) = n/d elements, so a^k generates G iff n/d = n, i.e. iff gcd(k, n) = 1.

By definition, the number of k with 1 ≤ k ≤ n and gcd(k, n) = 1 is φ(n), so G has exactly φ(n) generators.

Hence proved.

17. Division Algorithm in F[x]

Statement: Let F be a field and f, g ∈ F[x] with g ≠ 0. Then there exist unique q, r ∈ F[x] with f = qg + r, where r = 0 or deg r < deg g.

Proof: Existence. Let m = deg g. If f = 0 or deg f < m, take q = 0, r = f. Otherwise use induction on n = deg f ≥ m.

Write f = a_nxⁿ + ... and g = b_mx^m + ... with b_m ≠ 0. As F is a field, b_m⁻¹ exists; put f₁ = f − a_nb_m⁻¹x^n−mg.

The xⁿ terms cancel, so f₁ = 0 or deg f₁ < n. By the induction hypothesis (or trivially) f₁ = q₁g + r with r = 0 or deg r < m.

Then f = (a_nb_m⁻¹x^n−m + q₁)g + r, which has the required form.

Uniqueness. If f = qg + r = q′g + r′, then (q − q′)g = r′ − r. If q ≠ q′, the left side has degree ≥ deg g (degrees add in the integral domain F[x]), while r′ − r is 0 or of degree < deg g; this is impossible.

So q = q′, and then r = r′.

Hence proved.

1. Groups, Subgroups and Cosets

Problem 1 (10 marks): Show that G = {[[a, a], [a, a]] : a ∈ ℝ, a ≠ 0} is a group under matrix multiplication, although none of its elements is an invertible matrix.

Closure: [[a, a], [a, a]]·[[b, b], [b, b]] = [[2ab, 2ab], [2ab, 2ab]] ∈ G, since 2ab ≠ 0.

Associativity is inherited from matrix multiplication.

Identity: we need 2ae = a for every a ≠ 0, so e = 1/2. E = [[1/2, 1/2], [1/2, 1/2]] ∈ G and AE = EA = A.

Inverse: for A = [[a, a], [a, a]] we need 2ab = 1/2, so b = 1/(4a) ≠ 0, and [[b, b], [b, b]] ∈ G.

The product is symmetric in a and b, so G is abelian. Each element has determinant a² − a² = 0, so the group inverse is not a matrix inverse.

Answer: G is an abelian group with identity [[1/2, 1/2], [1/2, 1/2]]; the inverse of [[a, a], [a, a]] is [[1/(4a), 1/(4a)], [1/(4a), 1/(4a)]].

Problem 2 (15 marks): Find the order of every element of U(15), list all its subgroups, and decide whether U(15) is cyclic.

U(15) = {1, 2, 4, 7, 8, 11, 13, 14} under multiplication mod 15, of order φ(15) = 8.

4² = 16 ≡ 1, 11² = 121 ≡ 1 and 14² = 196 ≡ 1, so 4, 11, 14 have order 2.

2² = 4 and 2⁴ = 16 ≡ 1, so o(2) = 4; similarly 7² ≡ 8² ≡ 13² ≡ 4, so 2, 7, 8, 13 all have order 4.

No element has order 8, so U(15) is not cyclic.

Subgroups of order 2 are generated by elements of order 2: {1, 4}, {1, 11}, {1, 14}. Cyclic subgroups of order 4: ⟨2⟩ = ⟨8⟩ = {1, 2, 4, 8} and ⟨7⟩ = ⟨13⟩ = {1, 4, 7, 13}.

A non-cyclic subgroup of order 4 must consist of e and three elements of order 2: {1, 4, 11, 14}, which is closed (4·11 ≡ 14, 4·14 ≡ 11, 11·14 ≡ 4).

Adding {1} and U(15) itself gives all subgroups (orders must divide 8).

Answer: Orders: 1 → 1; 4, 11, 14 → 2; 2, 7, 8, 13 → 4. There are 8 subgroups: {1}, {1, 4}, {1, 11}, {1, 14}, {1, 2, 4, 8}, {1, 4, 7, 13}, {1, 4, 11, 14}, U(15). U(15) ≅ ℤ₂ × ℤ₄ is not cyclic.

Problem 3 (10 marks): Let a, b be elements of a group G with ab = ba, o(a) = m, o(b) = n and gcd(m, n) = 1. Prove that o(ab) = mn.

Since ab = ba, (ab)^mn = (a^m)ⁿ(bⁿ)^m = e, so o(ab) divides mn.

Let (ab)^k = e. Then a^kb^k = e, so a^k = b^−k ∈ ⟨a⟩ ∩ ⟨b⟩.

⟨a⟩ ∩ ⟨b⟩ is a subgroup of both ⟨a⟩ and ⟨b⟩, so by Lagrange its order divides both m and n, hence divides gcd(m, n) = 1. So ⟨a⟩ ∩ ⟨b⟩ = {e}.

Therefore a^k = e and b^k = e, giving m | k and n | k. As gcd(m, n) = 1, mn | k.

So the least such k is mn.

Answer: o(ab) = mn.

Problem 4 (15 marks): Let H = {e, (1 2)} in S₃. Find all left cosets and all right cosets of H, and decide whether H is normal in S₃. (Compose permutations right to left.)

[S₃ : H] = 6/2 = 3, so there are three left and three right cosets.

(1 3)(1 2) = (1 2 3) and (2 3)(1 2) = (1 3 2). So the left cosets are H, (1 3)H = {(1 3), (1 2 3)} and (2 3)H = {(2 3), (1 3 2)}.

(1 2)(1 3) = (1 3 2) and (1 2)(2 3) = (1 2 3). So the right cosets are H, H(1 3) = {(1 3), (1 3 2)} and H(2 3) = {(2 3), (1 2 3)}.

Since (1 3)H ≠ H(1 3), H is not normal. Equivalently, (1 3)(1 2)(1 3)⁻¹ = (3 2) ∉ H.

Answer: Left cosets: H, {(1 3), (1 2 3)}, {(2 3), (1 3 2)}. Right cosets: H, {(1 3), (1 3 2)}, {(2 3), (1 2 3)}. H is not normal in S₃.

2. Normal Subgroups, Homomorphisms and Permutation Groups

Problem 5 (10 marks): Prove that (ℤ × ℤ)/⟨(1, 1)⟩ ≅ ℤ.

Define f : ℤ × ℤ → ℤ by f(a, b) = a − b.

f((a, b) + (c, d)) = (a + c) − (b + d) = f(a, b) + f(c, d), so f is a homomorphism.

f is onto, since f(n, 0) = n for every n ∈ ℤ.

ker f = {(a, b) : a = b} = {a(1, 1) : a ∈ ℤ} = ⟨(1, 1)⟩.

By the first isomorphism theorem, (ℤ × ℤ)/⟨(1, 1)⟩ ≅ ℤ.

Answer: The map (a, b) + ⟨(1, 1)⟩ ↦ a − b is an isomorphism of (ℤ × ℤ)/⟨(1, 1)⟩ onto ℤ.

Problem 6 (15 marks): Prove that the centre of S_n is trivial for n ≥ 3. What is Z(S₂)?

Let σ ∈ S_n with σ ≠ e. Then σ(i) = j for some i ≠ j.

Since n ≥ 3, choose k ∉ {i, j} and let τ = (j k). Then τ⁻¹ = τ and τ(i) = i.

(τστ⁻¹)(i) = τ(σ(τ(i))) = τ(σ(i)) = τ(j) = k, whereas σ(i) = j ≠ k.

So τστ⁻¹ ≠ σ, i.e. τσ ≠ στ, and σ ∉ Z(S_n). Hence Z(S_n) = {e}.

For n = 2, S₂ = {e, (1 2)} is abelian, so Z(S₂) = S₂; the proof fails there because no third symbol k exists.

Answer: Z(S_n) = {e} for n ≥ 3, and Z(S₂) = S₂.

Problem 7 (10 marks): How many group homomorphisms are there from ℤ₂₀ to ℤ₁₀? How many of them are onto?

A homomorphism f is determined by a = f(1), with f(x) = xa (mod 10).

The rule is well defined iff 20a ≡ 0 (mod 10), which holds for every a ∈ ℤ₁₀. So there are 10 homomorphisms; in general the number from ℤ_m to ℤ_n is gcd(m, n) = gcd(20, 10) = 10.

f is onto iff its image ⟨a⟩ is all of ℤ₁₀, i.e. iff a is a generator: gcd(a, 10) = 1, so a ∈ {1, 3, 7, 9}.

Answer: 10 homomorphisms, of which 4 (f(1) = 1, 3, 7, 9) are onto.

Problem 8 (15 marks): Q₈ = {±1, ±i, ±j, ±k} with i² = j² = k² = −1, ij = k = −ji, jk = i = −kj, ki = j = −ik. Show that Q₈ is non-abelian but every subgroup of Q₈ is normal.

ij = k but ji = −k ≠ k, so Q₈ is non-abelian.

Orders: 1 has order 1, −1 has order 2, and ±i, ±j, ±k have order 4 (for example i² = −1, i⁴ = 1). So −1 is the only element of order 2.

Possible subgroup orders are 1, 2, 4, 8. Order 2: only {±1}. Order 4: a Klein group would need three elements of order 2, so each is cyclic: ⟨i⟩ = {±1, ±i}, ⟨j⟩ = {±1, ±j}, ⟨k⟩ = {±1, ±k}.

{1} and Q₈ are normal. {±1} is normal because ±1 commute with everything (it is the centre).

Each subgroup of order 4 has index 2, hence is normal.

Answer: Q₈ is non-abelian, and all six of its subgroups {1}, {±1}, ⟨i⟩, ⟨j⟩, ⟨k⟩, Q₈ are normal (Q₈ is a Hamiltonian group).

3. Rings, Ideals and Fields

Problem 9 (10 marks): In ℤ₁₂, find all units, zero divisors, idempotents (x² = x) and nilpotent elements.

Units are the k with gcd(k, 12) = 1: 1, 5, 7, 11. Each is its own inverse, since 25, 49 and 121 are all ≡ 1 (mod 12).

In a finite commutative ring every non-zero element is a unit or a zero divisor, so the zero divisors are 2, 3, 4, 6, 8, 9, 10 (for example 4·3 = 0, 9·4 = 0, 10·6 = 0).

Idempotents: x² ≡ x means 12 | x(x − 1). Checking, x = 0, 1, 4 (16 ≡ 4) and 9 (81 ≡ 9).

Nilpotents: xⁿ ≡ 0 requires 2 | x and 3 | x, so x = 0 or 6, and indeed 6² = 36 ≡ 0.

Answer: Units 1, 5, 7, 11; zero divisors 2, 3, 4, 6, 8, 9, 10; idempotents 0, 1, 4, 9; nilpotents 0, 6.

Problem 10 (10 marks): Show that ℤ[√2] = {a + b√2 : a, b ∈ ℤ} is an integral domain with infinitely many units.

ℤ[√2] contains 1, is closed under subtraction, and (a + b√2)(c + d√2) = (ac + 2bd) + (ad + bc)√2, so it is a subring of ℝ.

As a subring of the field ℝ containing 1, it is commutative with unity and has no zero divisors, so it is an integral domain.

(1 + √2)(−1 + √2) = 2 − 1 = 1, so u = 1 + √2 is a unit with inverse −1 + √2; hence every power uⁿ (n ∈ ℤ) is a unit.

Since u > 1, the numbers 1 < u < u² < u³ < ... are distinct, so there are infinitely many units.

Answer: ℤ[√2] is an integral domain, and ±(1 + √2)ⁿ, n ∈ ℤ, are infinitely many distinct units.

Problem 11 (15 marks): Let R = C[0, 1] be the ring of continuous real functions on [0, 1] under pointwise addition and multiplication. Show that M = {f ∈ R : f(1/2) = 0} is a maximal ideal of R. Is R an integral domain?

Define φ : R → ℝ by φ(f) = f(1/2). Then φ(f + g) = φ(f) + φ(g) and φ(fg) = φ(f)φ(g), so φ is a ring homomorphism.

φ is onto: for c ∈ ℝ the constant function c is continuous and φ(c) = c.

ker φ = {f : f(1/2) = 0} = M, so M is an ideal and, by the first isomorphism theorem, R/M ≅ ℝ.

ℝ is a field, so M is a maximal ideal.

R is not an integral domain: f(x) = max(0, x − 1/2) and g(x) = max(0, 1/2 − x) are continuous and non-zero, but f(x)g(x) = 0 for every x.

Answer: M is a maximal ideal since R/M ≅ ℝ; R is not an integral domain.

Problem 12 (15 marks): Show that F = ℚ[x]/⟨x³ − 2⟩ is a field, and find the inverse of 1 + α in F, where α = x + ⟨x³ − 2⟩.

p(x) = x³ − 2 has no rational root: by the rational root test the candidates are ±1, ±2, giving p = −1, −3, 6, −10.

A cubic with no root in ℚ has no linear factor, so p(x) is irreducible over ℚ.

ℚ[x] is a PID, so ⟨p(x)⟩ is a maximal ideal and F is a field. Its elements are a + bα + cα² (a, b, c ∈ ℚ) with α³ = 2.

Using a³ + b³ = (a + b)(a² − ab + b²): (1 + α)(1 − α + α²) = 1 + α³ = 1 + 2 = 3.

Hence (1 + α)⁻¹ = (1 − α + α²)/3.

Answer: F is a field (isomorphic to ℚ(∛2)), and (1 + α)⁻¹ = (1 − α + α²)/3.

Problem 13 (10 marks): Show that x⁴ + 1 is irreducible over ℚ.

Eisenstein's criterion: if f = a_nxⁿ + ... + a₀ ∈ ℤ[x] and a prime p divides a₀, ..., a_n−1, p does not divide a_n, and p² does not divide a₀, then f is irreducible over ℚ.

No prime works directly for x⁴ + 1, so substitute x = y + 1: f(y + 1) = y⁴ + 4y³ + 6y² + 4y + 2.

With p = 2: 2 divides 4, 6, 4, 2; 2 does not divide the leading coefficient 1; 4 does not divide 2. So f(y + 1) is irreducible over ℚ.

If f(x) = g(x)h(x) with deg g, deg h ≥ 1, then f(y + 1) = g(y + 1)h(y + 1) with the same degrees, a contradiction. So f is irreducible.

Answer: x⁴ + 1 is irreducible over ℚ (Eisenstein with p = 2 after the shift x = y + 1).

1. Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem

2025 Paper II, Q1(a), 10 marks: Let H and K be two subgroups of a group G such that o(H) > √o(G) and o(K) > √o(G). Show that H ∩ K ≠ {e}, where e is the identity element. Here o(H), o(K) and o(G) denote the order of H, K and G respectively.

2021 Paper II, Q4(a), 15 marks: Show that there are infinitely many subgroups of the additive group ℚ of rational numbers.

2020 Paper II, Q2(a), 15 marks: Let G be a finite cyclic group of order n. Then prove that G has φ(n) generators (where φ is Euler’s φ-function).

2019 Paper II, Q1(a), 10 marks: Let G be a finite group, H and K subgroups of G such that K ⊂ H. Show that (G : K) = (G : H)(H : K).

2018 Paper II, Q3(a), 20 marks: Find all the proper subgroups of the multiplicative group of the field (ℤ₁₃, +₁₃, ×₁₃), where +₁₃ and ×₁₃ represent addition modulo 13 and multiplication modulo 13 respectively.

2016 Paper II, Q2(b), 15 marks: Let p be a prime number and ℤ_p denote the additive group of integers modulo p. Show that every non-zero element of ℤ_p generates ℤ_p.

2015 Paper II, Q1(a), 10 marks: (i) How many generators are there of the cyclic group G of order 8? Explain.
(ii) Taking a group {e, a, b, c} of order 4, where e is the identity, construct composition tables showing that one is cyclic while the other is not. (5 + 5 = 10)

2. Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems

2024 Paper II, Q2(b), 15 marks: Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

2023 Paper II, Q1(a), 10 marks: Let G be a group of order 10 and G′ be a group of order 6. Examine whether there exists a homomorphism of G onto G′.

2022 Paper II, Q1(a), 10 marks: Show that the multiplicative group G = {1, −1, i, −i}, where i = √(−1), is isomorphic to the group G′ = ({0, 1, 2, 3}, +₄).

2022 Paper II, Q2(b), 15 marks: Prove that every homomorphic image of a group G is isomorphic to some quotient group of G.

2020 Paper II, Q1(a), 10 marks: Let S₃ and Z₃ be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S₃ in Z₃ except the trivial homomorphism.

2019 Paper II, Q2(a), 10 marks: If G and H are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from G to H, the trivial one.

2019 Paper II, Q2(b), 10 marks: Write down all quotient groups of the group Z₁₂.

2018 Paper II, Q2(a), 15 marks: Show that the quotient group of (ℝ, +) modulo ℤ is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here ℝ is the set of real numbers and ℤ is the set of integers.

2014 Paper II, Q1(a), 10 marks: Let G be the set of all real 2 × 2 matrices [[x, y], [0, z]], where xz ≠ 0. Show that G is a group under matrix multiplication. Let N denote the subset {[[1, a], [0, 1]] : a ∈ ℝ}. Is N a normal subgroup of G? Justify your answer.

3. Permutation Groups and Cayley's Theorem

2017 Paper II, Q1(b), 10 marks: Let G be a group of order n. Show that G is isomorphic to a subgroup of the permutation group S_n.

4. Direct Products, Dihedral Groups, Class Equation and Sylow Theorems

2026 Paper II, Q1(a), 10 marks: Let G be a group of order pq, where p and q are primes. Show that either G is an abelian group or no non-identity element commutes with every element of G.

2026 Paper II, Q2(a), 15 marks: Let G be a finite group generated by two elements u and v of order 2 each. Show that G is isomorphic to the dihedral group of order 2n, where O(uv) = n.

2025 Paper II, Q1(b), 10 marks: Let G = {e, x, x², y, yx, yx²} be a non-Abelian group with o(x) = 3 and o(y) = 2. Show that xy = yx² (where e is the identity element of G and o(x), o(y) denote the order of the elements x, y respectively).

2024 Paper II, Q1(a), 10 marks: Let G be a finite group of order mn, where m and n are prime numbers with m > n. Show that G has at most one subgroup of order m.

2023 Paper II, Q2(a), 15 marks: Prove that a non-commutative group of order 2p, where p is an odd prime, must have a subgroup of order p.

2017 Paper II, Q3(a), 15 marks: Show that the groups ℤ₅ × ℤ₇ and ℤ₃₅ are isomorphic.

5. Rings, Subrings, Ideals and Ring Homomorphisms

2025 Paper II, Q4(a), 15 marks: Examine whether the mapping φ : Z[x] → Z defined by φ(f(x)) = f(0), for f(x) ∈ Z[x], is a homomorphism. Deduce that the ideal ⟨x⟩ is a prime ideal in Z[x], but not a maximal ideal in Z[x].

2024 Paper II, Q4(a), 15 marks: Consider the polynomial ring Z[x] over the ring Z of integers. Let S be an ideal of Z[x] generated by x. Show that S is prime but not a maximal ideal of Z[x].

2022 Paper II, Q4(a), 15 marks: Let R be a field of real numbers and S, the field of all those polynomials f(x) ∈ R[x] such that f(0) = 0 = f(1). Prove that S is an ideal of R[x]. Is the residue class ring R[x]/S an integral domain? Give justification for your answer.

2018 Paper II, Q4(a), 20 marks: Suppose ℝ be the set of all real numbers and f : ℝ → ℝ is a function such that the following equations hold for all x, y ∈ ℝ :
(i) f(x + y) = f(x) + f(y)
(ii) f(xy) = f(x) f(y)
Show that ∀ x ∈ ℝ either f(x) = 0, or, f(x) = x.

2015 Paper II, Q1(b), 10 marks: Give an example of a ring having identity but a subring of this having a different identity.

2015 Paper II, Q2(a), 15 marks: If R is a ring with unit element 1 and φ is a homomorphism of R onto R′, prove that φ(1) is the unit element of R′.

2014 Paper II, Q4(a), 15 marks: Prove that the set ℚ(√5) = {a + b√5 : a, b ∈ ℚ} is a commutative ring with identity.

6. Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields

2026 Paper II, Q1(b), 10 marks: Show that every non-zero prime ideal in a Euclidean Domain is maximal.

2026 Paper II, Q4(a), 15 marks: Let n be a positive integer. Show that the following statements are equivalent:
(i) ℤ/nℤ is an integral domain.
(ii) ℤ/nℤ is a field.
(iii) n is prime.

2025 Paper II, Q2(b), 15 marks: Show that 3 is an irreducible element in the integral domain Z[i].

2023 Paper II, Q1(b), 10 marks: Express the ideal 4Z + 6Z as a principal ideal in the integral domain Z.

2023 Paper II, Q3(a), 15 marks: Prove that x² + 1 is an irreducible polynomial in Z₃[x]. Further show that the quotient ring Z₃[x]/⟨x² + 1⟩ is a field of 9 elements.

2021 Paper II, Q1(a), 10 marks: Let m₁, m₂, ⋯, m_k be positive integers and d > 0 the greatest common divisor of m₁, m₂, ⋯, m_k. Show that there exist integers x₁, x₂, ⋯, x_k such that d = x₁m₁ + x₂m₂ + ⋯ + x_km_k

2020 Paper II, Q1(b), 10 marks: Let R be a principal ideal domain. Show that every ideal of a quotient ring of R is principal ideal and R/P is a principal ideal domain for a prime ideal P of R.

2019 Paper II, Q3(d), 10 marks: Let a be an irreducible element of the Euclidean ring R, then prove that R/(a) is a field.

2018 Paper II, Q1(a), 10 marks: Let R be an integral domain with unit element. Show that any unit in R[x] is a unit in R.

2017 Paper II, Q2(c), 20 marks: Let F be a field and F[X] denote the ring of polynomials over F in a single variable X. For f(X), g(X) ∈ F[X] with g(X) ≠ 0, show that there exist q(X), r(X) ∈ F[X] such that degree (r(X)) < degree (g(X)) and
f(X) = q(X) . g(X) + r(X).

2016 Paper II, Q1(a), 10 marks: Let 𝕂 be a field and 𝕂[X] be the ring of polynomials over 𝕂 in a single variable X. For a polynomial f ∈ 𝕂[X], let (f) denote the ideal in 𝕂[X] generated by f. Show that (f) is a maximal ideal in 𝕂[X] if and only if f is an irreducible polynomial over 𝕂.

2015 Paper II, Q4(a), 15 marks: Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields:
(i) The set of numbers of the form b√2 with b rational
(ii) The set of even integers
(iii) The set of positive integers (5 + 6 + 4 = 15)

2014 Paper II, Q2(a), 15 marks: Show that ℤ₇ is a field. Then find ([5] + [6])⁻¹ and (−[4])⁻¹ in ℤ₇.

2014 Paper II, Q3(a), 15 marks: Show that the set {a + bω : ω³ = 1}, where a and b are real numbers, is a field with respect to usual addition and multiplication.

7. Field Extensions, Algebraic Elements and Finite Fields

2021 Paper II, Q2(b), 15 marks: Let F be a field and f(x) ∈ F[x] a polynomial of degree > 0 over F. Show that there is a field F′ and an imbedding q : F → F′ s.t. the polynomial f^q ∈ F′[x] has a root in F′, where f^q is obtained by replacing each coefficient a of f by q(a).

2020 Paper II, Q3(a), 15 marks: Let R be a finite field of characteristic p (> 0). Show that the mapping f : R → R defined by f(a) = a^p, ∀a ∈ R is an isomorphism.

2016 Paper II, Q3(a), 20 marks: Let K be an extension of a field F. Prove that the elements of K, which are algebraic over F, form a subfield of K. Further, if F ⊂ K ⊂ L are fields, L is algebraic over K and K is algebraic over F, then prove that L is algebraic over F.

2016 Paper II, Q4(a), 15 marks: Show that every algebraically closed field is infinite.