Groups

Lecture 13: Subgroup

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Lecture 08 - Subgroups

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lecture 14: Cyclic Group

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Lecture 16 - Cosets and Lagrange's theorem

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lagrange in full — and note that it gives no converse.

Lecture 16: Left Cosets

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Cosets partition the group — that is all Lagrange's theorem is.

Lagrange, orders and cyclic groups free

In plain words

A subgroup's size always divides the group's size. The reverse fails: a divisor need not appear as a subgroup, and $A_4$ has twelve elements and no subgroup of six.

Lagrange and its partial converses

|H| divides |G|, and o(g) divides |G|, so $g^{|G|}$ = e.

Statement	True?
d \| \|G\| ⇒ G has a subgroup of order d	✗ — $A_{4}$ has order 12, no subgroup of order 6
p prime, p \| \|G\| ⇒ G has an element of order p	✓ Cauchy
pᵏ \| \|G\| ⇒ G has a subgroup of order pᵏ	✓ Sylow
G abelian and d \| \|G\| ⇒ subgroup of order d	✓
G cyclic ⇒ exactly one subgroup of each order dividing \|G\|	✓

Cyclic groups

$\mathbb{Z}_{n}$ has $\varphi(d)$ elements of order d for each d | n, and exactly one subgroup of each such order. |Aut $(\mathbb{Z}_{n})$ | $= \varphi(n)$ .

Union of proper subgroups: a finite group is the union of its proper subgroups iff it is not cyclic — a generator lies in no proper subgroup, and conversely every element generates a proper one. No group over any field is the union of two proper subgroups.

Order arithmetic

o(ab) = o(a)o(b) when a, b commute and gcd(o(a), o(b)) = 1. Without commuting it can be infinite: in GL $_{2}$ two elements of order 2 can multiply to infinite order.
Index 2 ⇒ normal. Index p (smallest prime dividing |G|) ⇒ normal.
|G| = p ⇒ cyclic. |G| $= p^{2} \Rightarrow$ abelian $(\mathbb{Z}_{p^{2}}$ or $\mathbb{Z}_p \times \mathbb{Z}_p)$ , not necessarily cyclic.
|G| = pq with p < q and p ∤ q − 1 ⇒ cyclic.

Key takeaways

Converse of Lagrange fails $(A_{4}$ , order 6); Cauchy and Sylow are the true partial converses.
Non-cyclic ⇔ union of proper subgroups. $- p^{2} \Rightarrow$ abelian, but not necessarily cyclic.

Lagrange's theorem, tiledinteractive

This visual is part of Trap Analysis

Cosets colouring a cyclic group: same size, never overlapping, always tiling it exactly. The picture will not let you build a counterexample.

Lagrange's theorem and its false converseinteractive

This visual is part of Trap Analysis

Order divides — but a divisor need not give a subgroup, and A₄ is the reason.

Exam focus: Lagrange gives one direction only. The exam tests the failures: no subgroup of a given order (A₄), and 'union of proper subgroups ⇔ not cyclic'.

Practice

Part B2025 Decembershow ▾

For a finite group G, let H_G = {g ∈ G ∣

g^{15} = e

}. Which of the following statements is necessarily true?

A.There exists a finite group G such that |H_G| is even.
B.There exists a finite group G such that |H_G| = 4n+1 for some n ≥ 3.✓ Correct
C.For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+1.
D.For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+3.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

For a finite group G, let S(G) denote the number of subgroups of G. Which of the following statements is necessarily true?

A.Let G and G′ be finite groups such that S(G) = S(G′). Then G is isomorphic to G′.
B.If S(G) = 4, then |G| = pᵐ for some prime number p and positive integer m.
C.If S(G) = 5, then G is a cyclic group.
D.For every positive integer n, there exists a finite group G such that S(G) = n.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

Which of the following statements is true?

A.p ∤ 1 + (p − 1)! for some odd prime p.
B.p | $(1234)^{p-1}$ − 1 for all primes p > 700.✓ Correct
C.There exist $a \in \mathbb{Z}$ and a prime p > 11 such that p ∤ aᵖ − a.
D.p ∤ $(p^{2})!/(p!)^{2}$ for some odd prime p.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let p be a prime number. An element a of the multiplicative group

(\mathbb{Z}/p\mathbb{Z})

ˣ is said to be a primitive root in

(\mathbb{Z}/p\mathbb{Z})

ˣ if the order of a in

(\mathbb{Z}/p\mathbb{Z})

ˣ is p − 1. Let S_p be the number of primitive roots in

(\mathbb{Z}/p\mathbb{Z})

ˣ and

\varphi

denote the Euler

\varphi-

function. Which of the following statements is true?

A.For each $p < 100, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ converges.
B.For each $100 \le p \le 200, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ diverges.
C.For each $p > 200, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ converges.✓ Correct
D.The element 4 mod 101 in $(\mathbb{Z}/101\mathbb{Z})$ ˣ is a primitive root.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

For integers n > 1, let G(n) denote the number of groups of order n, up to isomorphism, i.e. G(n) is the number of isomorphism classes of groups of order n. Which of the following statements is true?

A.If G(n) = 1, then n is prime.
B.G(8) = 2
C.If $\gcd(n, \varphi(n)) > 1$ , then G(n) > 1. (Here $\varphi$ denotes the Euler $\varphi-$ function.)✓ Correct
D. $\lim \sup(n\to\infty) G(n) = 2$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let G be a finite group. For any prime number p, let E(p) = {g ∈ G | g^p = 1}. Which of the following statements are true?

A.If p divides |G|, then E(p) is a subgroup of G.
B.For all p, E(p) is a subgroup of G.
C.If E(p) is a subgroup of G, then p divides |G|.
D.If G is abelian, then E(p) is a subgroup of G.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

The number of group homomorphisms from

\mathbb{Z}/150\mathbb{Z}

\mathbb{Z}/90\mathbb{Z}

A.30✓ Correct
B.60
C.45
D.10

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Which of the following statements are true?

A.Let $G_{1}$ and $G_{2}$ be finite groups such that | $G_{1}$ | and | $G_{2}$ | are coprime. Then any homomorphism from $G_{1}$ to $G_{2}$ is trivial.✓ Correct
B.Let G be a finite group and f : G → G a homomorphism that fixes more than half of the elements of G. Then f(x) = x for all x ∈ G.✓ Correct
C.Let G be a finite group having exactly 3 subgroups. Then G is of order $p^{2}$ for some prime p.✓ Correct
D.Any finite abelian group G has at least d(|G|) subgroups, where d(m) is the number of positive divisors of m.✓ Correct

Solution

(1) the image has order dividing both.

(2) the fixed set is a subgroup of index < 2, hence all of G.

(3) exactly three subgroups forces cyclic of order $p^{2}$ .

(4) an abelian group has a subgroup of every order dividing |G|.

Part B2023 Decembershow ▾

Let G be any finite group. Which one of the following is necessarily true?

A.G is a union of proper subgroups.
B.G is a union of proper subgroups if |G| has at least two distinct prime divisors.
C.If G is abelian, then G is a union of proper subgroups.
D.G is a union of proper subgroups if and only if G is not cyclic.✓ Correct

Solution

A finite group is the union of its proper subgroups iff it is not cyclic: if it is cyclic a generator lies in no proper subgroup; if not cyclic every element generates a proper subgroup.

Part Bshow ▾

The number of generators of the cyclic group

\mathbb{Z}/12\mathbb{Z}

A.2
B.4✓ Correct
C.6
D.12

Solution

Generators of $\mathbb{Z}/n$ are the residues coprime to n, counted by Euler $'s \varphi: \varphi(12) = 12(1 - 1/2)(1 - 1/3) = 4$ , namely 1, 5, 7, 11.

Part Cshow ▾

Let G be a finite group and H ≤ G. Which of the following are true?

A.|H| divides |G| (Lagrange's theorem).✓ Correct
B.The left cosets of H partition G into |G|/|H| classes, each of size |H|.✓ Correct
C.If d divides |G|, then G has a subgroup of order d.
D.Every subgroup of a cyclic group is cyclic.✓ Correct

Solution

(1)–(2) are Lagrange's theorem and its immediate consequence. (4) is a standard fact about cyclic groups. (3) is false — the converse of Lagrange's theorem fails in general: $A_{4}$ has order 12, and 6 divides 12, but $A_{4}$ has no subgroup of order 6.

Part Cshow ▾

Which of the following statements about cyclic groups are true?

A.Every subgroup of a cyclic group is cyclic.✓ Correct
B.A cyclic group of order n has exactly one subgroup of each order dividing n.✓ Correct
C. $(\mathbb{Q}, +)$ is cyclic.
D.Every group of prime order is cyclic.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

G = \mathbb{Z}/12\mathbb{Z}

. Which of the following are true?

A.G has exactly one subgroup of each order dividing 12.✓ Correct
B.The element 5 generates G.✓ Correct
C.The number of generators of G is $\varphi(12) = 4$ .✓ Correct
D.Every subgroup of G has a unique complementary subgroup K, with H ∩ K = {0} and H + K = G.

Solution

(1) is the standard fact that a cyclic group has exactly one subgroup for each divisor of its order. (2) gcd(5, 12) = 1, so 5 generates $G. (3) \varphi(12) = \varphi(4)\varphi(3) = 2\cdot2 = 4. (4)$ is false: the unique subgroup of order 6 is ⟨2⟩ = {0,2,4,6,8,10}, which already contains ⟨6⟩ = {0, 6} — so ⟨6⟩ has no complement of order 6 at all, let alone a unique one.

Part Bshow ▾

The number of group homomorphisms from

\mathbb{Z}_{6}

\mathbb{Z}_{4}

A.1
B.2✓ Correct
C.4
D.6

The worked solution and the per-option analysis are in Trap Analysis.

2. Normal subgroups, quotients, isomorphism theorems

Lecture 18: Normal Subgroup

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Normality is what makes the quotient a group; it is not transitive.

Lecture 13 - Normal subgroups

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lecture 19 - Quotient groups

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lecture 21 - First isomorphism theorem

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

The isomorphism theorems, with worked examples in the next lecture.

Normal subgroups, quotients and the isomorphism theorems free

In plain words

A normal subgroup is one you are allowed to divide by. Say normal in something, because it is a relationship rather than a badge — which is exactly why normality does not chain.

When is H normal?

index 2 ⇒ normal; index = smallest prime dividing |G| ⇒ normal.
kernels are normal; centres and commutator subgroups are normal (indeed characteristic).
Every subgroup of an abelian group is normal. The converse fails: $Q_{8}$ is non-abelian with every subgroup normal (a Hamiltonian group).

Normality is not transitive: ⟨(12)(34)⟩ ⊴ $V_{4}$ ⊴ $A_{4}$ , but ⟨(12)(34)⟩ is not normal in $A_{4}$ . This is a recurring Part-C item.

Isomorphism theorems

$1. G/\ker \varphi \cong$ im $\varphi$ . 2. HN/N ≅ H/(H ∩ N). 3. (G/N)/(M/N) ≅ G/M for N ⊆ M ⊴ G. 4. Correspondence: subgroups of G/N ↔ subgroups of G containing N.

Simple groups

$- \mathbb{Z}_p$ and $A_{n}$ for n ≥ 5 are simple; $A_{5} ($ order 60) is the smallest non-abelian simple group.

A group of order pᵏ (k ≥ 2), pq, $p^{2}q$ , or any group with a normal Sylow subgroup, is not simple.
|G| $= 2023 = 7\cdot17^{2}$ : both Sylow counts are 1, so it is abelian and not simple.

Standard structures

G	normal subgroups
$S_{3}$	$1, A_{3}, S_{3}$
$S_{4}$	$1, V_{4}, A_{4}, S_{4}$
$A_{4}$	$1, V_{4}, A_{4} ($ no subgroup of order 6)
$Q_{8}$	all subgroups
p-group of order $p^{n}$	one of every order pᵏ, k ≤ n — so at least n + 1

Key takeaways

Index 2 ⇒ normal; normality does not pass through a chain.
Every subgroup normal ⇏ abelian $(Q_{8})$ .
A p-group of order $p^{n}$ has normal subgroups of every order pᵏ — count them before answering "exactly k normal subgroups" questions.

The first isomorphism theorem, used as a toolinteractive

This visual is part of Trap Analysis

Three quotients identified in one line each — the fastest way to answer 'what is G/N?'

Exam focus: Normality is not transitive. Know the standard normal/non-normal examples in S₄ and A₄, and that index-2 subgroups are always normal.

Practice

Part C2025 Juneshow ▾

For a group G, let Aut(G) denote the group (under composition) of all bijective group homomorphisms from G onto itself. Which of the following statements are true?

A.If $G_{1}, G_{2}$ are two groups such that Aut $(G_{1})$ is isomorphic to Aut $(G_{2})$ , then $G_{1}$ is isomorphic to $G_{2}$ .
B.If |G| = 2, then Aut(G × G) is abelian.
C.If G is the group of complex numbers under addition, then Aut(G) is abelian.
D.If G is finite, then Aut(G) is finite.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Let

G_{1}

and

G_{2}

be subgroups of a group G. Which of the following statements are true?

A.If $G_{1}$ is normal in G, then $(G_{2}G_{1})/G_{1} \cong G_{2}/(G_{2} \cap G_{1})$ .✓ Correct
B.If $H_{1}$ and $H_{2}$ are normal subgroups of $G_{1}$ and $G_{2}$ , respectively, then $(G_{1} \times G_{2})/(H_{1} \times H_{2}) \cong (G_{1}/H_{1}) \times (G_{2}/H_{2})$ .✓ Correct
C.If $G_{1}$ is normal in $G_{2}$ and $G_{2}$ is normal in G, then $G_{1}$ is normal in G.
D.Every subgroup of prime index in G is normal.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Let G be a finite non-abelian group. Which of the following statements are necessarily true?

A.If d is a positive integer that divides |G|, then G has a subgroup of order d.
B.The map f : G×G → G given by f(a,b)=ab is not a group homomorphism.✓ Correct
C.Suppose that for every positive integer d that divides |G|, there exists a subgroup of G of order d. Then G has at least three normal subgroups.✓ Correct
D.|G| ≠ 16

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

We say that a group G has property (A) if every non-trivial homomorphism from G to any group is injective. Which of the following groups has property (A)?

A.The cyclic group of order 6.
B.The symmetric group $S_{5}$ .
C.The alternating group $A_{5}$ .✓ Correct
D.The dihedral group with ten elements.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following statements are true?

A.Every subgroup of index 2 is normal✓ Correct
B.If every subgroup of G is normal then G is abelian
C.Normality is transitive: H ⊴ K ⊴ G implies H ⊴ G
D.The commutator subgroup [G, G] is normal in G✓ Correct

Solution

(b) fails for $Q_{8}$ .

Part Bshow ▾

The number of normal subgroups of a group of order

p^{5} (p

prime) is at least

A.2
B.4
C.6✓ Correct
D.p

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following are true?

A.Every subgroup of index 2 is normal✓ Correct
B.Every subgroup of an abelian group is normal✓ Correct
C. $A_{4}$ has a normal subgroup of order 4✓ Correct
D. $A_{4}$ has a subgroup of order 6

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

Let H be a subgroup of index 2 in a group G. Which statement is correct?

A.H is normal in G.✓ Correct
B.H is normal only if G is abelian.
C.H is normal only if G is finite.
D.H need not be normal.

The worked solution and the per-option analysis are in Trap Analysis.

3. Permutation groups: cycles, sign, conjugacy in S_n and A_n

Lecture 08: Permutation

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Cycle notation and sign — then use the counting formula in the notes.

Lecture 26 - Symmetric groups I

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Cycle notation and disjoint cycle decomposition.

Lecture 30 - Odd and even permutations I

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

The sign homomorphism — well-definedness is the subtle point.

Lecture 32 - Alternating groups

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Permutation groups — cycle type, sign and conjugacy free

In plain words

Break a shuffle into disjoint cycles and almost everything follows. The order is the lowest common multiple of the lengths, and a cycle of even length is an odd permutation.

Basics

Every permutation is a product of disjoint cycles; the order is the lcm of the cycle lengths.
A k-cycle is a product of k − 1 transpositions, so its sign is $(-1)^{k-1}$ . Even permutations form $A_{n}$ , index 2.
Conjugacy in $S_{n} \Leftrightarrow$ same cycle type. The number of conjugacy classes of $S_{n}$ is p(n), the number of partitions of n.

Counting

Number of permutations in $S_{n}$ with cycle type $1^{a_{1}}2^{a_{2}}\cdots: n! / (\prod$ $k^{a_k}$ a_k!).

Examples in $S_{5} ($ order 120): 1 identity, 10 transpositions, 20 3-cycles, 30 4-cycles, 24 5-cycles, 15 (2,2), 20 (3,2).

Elements of order 6 in $S_{5}$ : cycle type (3,2) — 20 of them. Order 4: 4-cycles — 30.

$A_{n}$

| $A_{n}$ | $= n!/2; A_{n}$ is simple for $n \ge 5; A_{4}$ has no subgroup of order 6.
A conjugacy class of $S_{n}$ contained in $A_{n}$ splits into two $A_{n}-$ classes iff the cycle type has distinct odd parts. In $A_{5}$ the twenty-four 5-cycles split into two classes of 12 — that is why $A_{5}$ has class sizes 1, 15, 20, 12, 12.

Generators

$S_{n} =$ ⟨(12), (12⋯n)⟩ = ⟨all transpositions⟩ = ⟨adjacent transpositions⟩. $A_{n}$ is generated by 3-cycles.

Cayley

Every group of order n embeds in $S_{n}$ . So questions of the form "is G isomorphic to a subgroup of $S_{k}$ " reduce to finding a suitable action.

Key takeaways

Order = lcm of cycle lengths; sign = $(-1)^{n - c}$ , where c is the number of cycles, counting fixed points.
Same cycle type ⇔ conjugate in $S_{n}$ ; in $A_{n}$ classes may split.
Use the $n!/(\prod{}k$ ᵃ $^{k}a_{k}!)$ formula for all counting questions.

Cycle type decides everything in Sₙinteractive

This visual is part of Trap Analysis

Order, sign and conjugacy all read off one decomposition — plus where the rule changes in Aₙ.

Exam focus: Cycle type determines conjugacy in Sₙ (but splits in Aₙ). Count elements of a given order by counting cycle types; know the order of a permutation is the lcm of its cycle lengths.

Practice

Part B2024 Juneshow ▾

How many arrangements of the digits of the number 1234567 are there, such that exactly three of them occur in their original position? (E.g., in the arrangement 5214763, exactly the digits 2, 4 and 6 are in their original positions. In the arrangement 1243576, exactly the digits 1, 2 and 5 are in their original positions.)

A.525
B.35
C.840
D.315✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

What is the number of injective functions from {1, 2, …, 7} to {1, 2, …, 10}?

A. $10^{7}$
B.10!/7!
C.10!/3!✓ Correct
D. $7^{10}$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Which of the following numbers are order of some element of the symmetric group

S_{5}

A.3✓ Correct
B.4✓ Correct
C.5✓ Correct
D.6✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

G = A_{4}

, the alternating group on 4 letters. Which of the following are true?

A. $A_{4}$ has no subgroup of order 6.✓ Correct
B. $A_{4}$ has a normal subgroup of order 4, the Klein four-group $V_{4}$ .✓ Correct
C. $A_{4}/V_{4} \cong \mathbb{Z}_{3}$ .✓ Correct
D. $A_{4}$ is simple (has no nontrivial proper normal subgroup).

Solution

(1) is the classic converse-of-Lagrange counterexample. $(2) V_{4} =$ {e, (12)(34), (13)(24), (14)(23)} is normal in $A_{4}. (3)$ | $A_{4}$ |/| $V_{4}$ | = 12/4 = 3, and every group of order 3 is cyclic. (4) is false, and directly contradicted by $(2): V_{4}$ is itself a nontrivial proper normal subgroup, so $A_{4}$ is not simple — simplicity only starts at $A_{5}$ .

Part Cshow ▾

Let

\sigma = (1 2 3)(4 5) \in S_{5} (

disjoint cycles). Which of the following are true?

A.The order of $\sigma$ is 6.✓ Correct
B. $\sigma$ is an odd permutation.✓ Correct
C. $\sigma$ is conjugate in $S_{5}$ to (1 3 5)(2 4).✓ Correct
D. $\sigma \in A_{5}$ .

Solution

(1) The order of a product of disjoint cycles is the lcm of the lengths: lcm(3, 2) = 6.

(2) A 3-cycle is even, a transposition odd, so the product is odd.

(3) Conjugacy in $S_{n}$ is exactly equality of cycle type, and both have type 3 + 2.

(4) fails: $A_{5}$ is the even permutations.

Part Bshow ▾

The number of elements of order 6 in

S_{5}

A.0
B.20✓ Correct
C.24
D.30

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

G = S_{6}

and

\sigma = (1 2)(3 4 5)

. Which of the following are true?

A. $\sigma$ has order 6.✓ Correct
B. $\sigma$ is an odd permutation.✓ Correct
C.The conjugacy class of $\sigma$ in $S_{6}$ has exactly 120 elements.✓ Correct
D. $\sigma$ commutes with every element of $A_{6}$ .

Solution

$\sigma$ has cycle type (2, 3, 1) — a 2-cycle, a 3-cycle, and a fixed point. (1) order lcm(2,3) = 6. (2) a 2-cycle is odd, a 3-cycle even, so $\sigma$ is odd. (3) the number of permutations of this cycle type is 6!/(2·3·1) = 120, since the cycle lengths are all distinct. (4) is false: by orbit–stabilizer the centralizer of $\sigma$ has order | $S_{6}$ |/120 = 6 — a tiny subgroup, nowhere near all of $A_{6} ($ order 360).

Part Cshow ▾

Which of the following are true in

S_{n}

A.Two permutations are conjugate iff they have the same cycle type✓ Correct
B.The order of a permutation is the lcm of its cycle lengths✓ Correct
C.Every permutation is a product of transpositions✓ Correct
D. $A_{n}$ is simple for all n ≥ 3

Solution

$A_{3} \cong \mathbb{Z}_{3}$ is simple but $A_{4}$ is not $(V_{4}$ is normal); simplicity holds for n ≥ 5.

Part Cshow ▾

Consider

S_{5}

and

A_{5}

. Which of the following are true?

A.Every element of $S_{5}$ can be written as a product of disjoint cycles.✓ Correct
B.A permutation of cycle type (3, 2) — a 3-cycle and a disjoint 2-cycle — is odd.✓ Correct
C.Two permutations in $S_{n}$ are conjugate if and only if they have the same cycle type.✓ Correct
D. $A_{5}$ contains an element of order 6.

Solution

(1) and (3) are the standard structural facts about $S_{n}$ .

(2) a 3-cycle is even (two transpositions) and a 2-cycle is odd (one transposition), so the product is odd.

(4) is false: | $A_{5}$ | = 60 is divisible by 6, but divisibility only bounds possible element orders via Lagrange, it doesn't produce them.

The cycle types of even permutations in $S_{5}$ are $1^{5}, 2^{2}1, 31^{2}$ , and 5, giving element orders 1, 2, 3, 5 only — no element of $A_{5}$ has order 6.

Part Cshow ▾

Let

\sigma = (1 2 3)(4 5) \in S_{5}

. Which of the following are true?

A. $\sigma$ has order 6.✓ Correct
B. $\sigma$ is an odd permutation.✓ Correct
C. $\sigma^{3} = (4 5)$ .✓ Correct
D. $\sigma \in A_{5}$ , the alternating group.

Solution

$\sigma$ is a disjoint product of a 3-cycle and a 2-cycle, so its order is lcm(3, 2) = 6 — (1) holds.

Its sign is the product of an even permutation (the 3-cycle) and an odd one (the 2-cycle), so $\sigma$ is odd — (2) holds. $\sigma^{3} = (1 2 3)^{3}(4 5)^{3} = e\cdot(4 5) = (4 5)$ — (3) holds.

(4) is false precisely because $\sigma$ is odd: $A_{5}$ consists only of even permutations, and an odd permutation can never belong to it.

4. Group actions, class equation, p-groups

Lecture 33 - Group actions

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lecture 35 - Orbits and stabilizers

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Orbit–stabiliser: every orbit size divides the group order.

Lecture 36 - Counting formula

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Lecture 39 - Problems 8 and Class equation

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

The class equation — this is what proves p-groups have a centre.

Group actions, the class equation and p-groups free

In plain words

A group acting on a set splits it into orbits whose sizes divide the group's size. That one counting fact carries the class equation and everything built on top of it.

Orbit–stabiliser and the class equation

|orbit(x)| = [G : Stab(x)], so every orbit size divides |G|.

Class equation: |G| = |Z(G)| $+ \sum [G : C(x_{i})]$ over representatives of the non-central classes.

Consequences

A non-trivial p-group has non-trivial centre (all non-central class sizes are divisible by p).
|G| $= p^{2} \Rightarrow G/Z$ is cyclic ⇒ G abelian.
If G/Z(G) is cyclic then G is abelian. Hence |G/Z| is never prime.
A p-group of order $p^{n}$ has a normal subgroup of each order pᵏ, and a chain of them.
Burnside: the number of orbits = (1/|G| $) \sum_g$ |Fix(g)| — used for colouring-count questions.

Standard actions

Action	Consequence
G on itself by left multiplication	Cayley: G ↪ $S_{\\|G\\|}$
G on itself by conjugation	class equation
G on cosets of H	G → $S_{[G:H]}$ ; kernel ⊆ H, so a subgroup of small index gives a normal subgroup
G on Sylow p-subgroups	Sylow's counting theorems

Index trick: if [G : H] = n and |G| ∤ n!, then H contains a non-trivial normal subgroup of G. This is how "no simple group of order 24/36/48" arguments go.

Key takeaways

p-groups always have a centre; that plus induction gives the whole normal-subgroup chain.
G/Z cyclic ⇒ abelian is the fastest way to kill many hypotheses.
The coset action turns "small index" into "not simple".

The class equation, and what it proves about p-groupsinteractive

This visual is part of Trap Analysis

Counting orbits mod p — the argument behind almost every small-group structure result.

Exam focus: The class equation is the engine: it proves p-groups have non-trivial centre, and drives the 'no simple group of order n' arguments.

Practice

Part C2025 Juneshow ▾

Let G be a group, H a subgroup of G, and T = {gH | g ∈ G}, the set of all left cosets of H in G. Let S_T be the set of all permutations of T and

\pi : G \to S_T

be the map defined by

\pi(g)(g_{1}H) =

_{1}H

. For a prime number p, let 𝔽_p denote the field with p elements. In which of the following cases is

\ker \pi

trivial?

A.G = GL $_{2}($ 𝔽_p) and H is a subgroup of order p.✓ Correct
B.G = SL $_{2}($ 𝔽_p) and H is a subgroup of order p.✓ Correct
C.p ≡ 3 (mod 4), G = GL $_{2}($ 𝔽_p)/SL $_{2}($ 𝔽_p) and H is a subgroup of order 2.
D.p ≡ 1 (mod 4), G = GL $_{2}($ 𝔽_p)/SL $_{2}($ 𝔽_p) and H is a subgroup of order 2.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

Let X = {1, …, 17} and

S_{17}

be the group of permutations of X. For a subgroup G of

S_{17}

and x ∈ X, let S_G(x) = {

\sigma \in G

∣

\sigma(x) = x

}. Which of the following statements is true for every subgroup G of

S_{17}

A.The number of pairs $(\sigma, x) \in G \times X$ such that $\sigma(x) = x$ is strictly greater than $\sum_{x}$ |S_G(x)|.
B.The number (1/|G| $)\cdot\sum_{x}$ |S_G(x)| is always an integer.✓ Correct
C.For all x ∈ X, S_G(x) is a normal subgroup of G.
D.For all x, y ∈ X, S_G(x) is isomorphic to S_G(y).

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let G be a group of order n > 3 and H be a subgroup with 1 < |H| < n. Consider the set X = ⋃(g∈G) gHg⁻

^{1}

. Which of the following statements is true?

A.If G is abelian, then |X| = n.
B.If |X| divides n, then G is abelian.
C.|X| < n✓ Correct
D.|X| divides n.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

Let p be a prime number. Let G be a group such that for each g ∈ G there exists an

n \in \mathbb{N}

such that

g^{p^{n}}

= 1. Which of the following statements is FALSE?

A.If |G| $= p^{6}$ , then G has a subgroup of index $p^{2}$ .
B.If |G| $= p^{6}$ , then G has at least five normal subgroups.
C.Center of G can be infinite.
D.There exists G with |G| $= p^{6}$ such that G has exactly six normal subgroups.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let G be the group (under matrix multiplication) of 2 × 2 invertible matrices with entries from

\mathbb{Z}/9\mathbb{Z}

. Let a be the order of G. Which of the following statements are true?

A.a is divisible by $3^{4}$ .✓ Correct
B.a is divisible by $2^{4}$ .✓ Correct
C.a is not divisible by 48.
D.a is divisible by $3^{6}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let G be a finite p-group, |G|

= p^{n}

with n ≥ 1. Which of the following are true?

A.The centre Z(G) is non-trivial.✓ Correct
B.If |G| $= p^{2}$ , then G is abelian.✓ Correct
C.Every p-group is cyclic.
D.G has a normal subgroup of order p.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let G be a group of order

p^{2}

for a prime p. Which of the following are true?

A.G is abelian.✓ Correct
B.G is isomorphic to $\mathbb{Z}/p^{2}\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ .✓ Correct
C.The center Z(G) of any nontrivial finite p-group is trivial.
D.|G| = |Z(G)| $+ \sum[G : C(x_{i})]$ , summed over representatives of the non-central conjugacy classes (the class equation).✓ Correct

Solution

(1)–(2) are the standard order $-p^{2}$ classification, proved using the class equation. (4) is the class equation itself. (3) is exactly backwards: the key lemma is that the center of a nontrivial finite p-group is always nontrivial — it is what makes the order $-p^{2}$ argument work in the first place.

Part Cshow ▾

Let G act on a finite set X. Which of the following are true?

A.By the Orbit–Stabilizer theorem, |Orbit(x)| = [G : Stab(x)] for every x ∈ X.✓ Correct
B.By Burnside's lemma, the number of orbits equals (1/|G| $)\sum_{g\in{}G}$ |Fix(g)|.✓ Correct
C.If G acts transitively on X, the stabilizers of different points are all conjugate subgroups of G.✓ Correct
D.If |G| = 20 and G acts on a set of size 3, the action must be trivial.

Solution

(1)–(3) are standard facts about group actions. (4) is false: orbit sizes must divide |G| = 20, and orbits of size 1, 2, and 3 items must divide 3 — only 1 and 2 do, so the possible orbit-size partitions of {1,2,3} are 1+1+1 or 1+2, forcing at least one fixed point, but not that every point is fixed. A partition like 1+2 has a genuine orbit of size 2, so the action need not be trivial.

5. Sylow theorems and groups of small order

Lecture 41 - Sylow Theorem I

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Existence of Sylow subgroups.

Lecture 42 - Sylow Theorem II

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Conjugacy of Sylow subgroups.

Lecture 43 - Sylow Theorem III

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

The counting theorem n_p ≡ 1 (mod p), n_p | m — the one you actually use.

Sylow theorems and classifying small groups free

In plain words

Sylow lets you describe a group knowing only how big it is. The number of subgroups of prime-power size divides the rest and leaves remainder one, and a count of one means normal.

The three theorems

Let |G| = pᵏm with p ∤ m.

Existence: Sylow p-subgroups (order pᵏ) exist.
Conjugacy: all are conjugate; every p-subgroup lies in one.
Counting: n_p ≡ 1 (mod p) and n_p | m. Also n_p = [G : N(P)].

n_p = 1 ⇔ the Sylow p-subgroup is normal.

The standard drill

\|G\|	Conclusion
pq, p < q, p ∤ q−1	cyclic
pq, p \| q−1	$\mathbb{Z}_{pq}$ or the non-abelian semidirect product
$p^{2}$	abelian: $\mathbb{Z}_{p^{2}}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$
15 = 3·5	$n_{3} = n_{5} = 1 \Rightarrow$ cyclic
20 = 4·5	$n_{5} \equiv 1 (mod 5), n_{5}$ \| $4 \Rightarrow n_{5} = 1$
$2023 = 7\cdot17^{2}$	$n_{7} = n_{17} = 1 \Rightarrow$ abelian, not simple
30 = 2·3·5	$n_{3} = 1$ or $n_{5} = 1 ($ counting elements) ⇒ not simple
36, 48, 24	use the coset action: index too small ⇒ not simple

Counting-elements argument

If n_q = k > 1 with q prime, the Sylow q-subgroups contribute k(q − 1) elements of order q. Adding up over primes and exceeding |G| gives a contradiction — the standard proof that no group of order 30 is simple.

Normal Sylow ⇒ direct product

If every Sylow subgroup is normal, G is the direct product of them (a nilpotent group). Finite abelian groups and p-groups are the examples.

Key takeaways

n_p ≡ 1 (mod p) and n_p | m — apply to each prime in turn.
One normal Sylow subgroup kills simplicity immediately.
All Sylow subgroups normal ⇒ G is their direct product.

Sylow counting: proving a group is not simpleinteractive

This visual is part of Trap Analysis

The standard argument, run on order 12 and order 15 — including the element count that finishes it.

Exam focus: n_p ≡ 1 (mod p) and n_p | m is the whole toolkit. Use it to force a normal Sylow subgroup and prove non-simplicity, or to classify groups of small order.

Practice

Part C2025 Decembershow ▾

Let G be a group of order 8. Which of the following statements are necessarily true?

A.If there are no elements of order 4 in G, then G is abelian.✓ Correct
B.If there are exactly two elements of order 4 in G, then G is abelian.
C.If there are exactly six elements of order 4 in G, then G is abelian.
D.There is an element of order 4 in G.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

For a finite group G, let S(G) denote the set of all its Sylow subgroups. A finite group G is said to have property (J) if G is isomorphic to the direct product

\prod(H\in{}S(G)) H

. Which of the following statements are true?

A.Any group with p(p + 2) elements, where p and p + 2 are both prime numbers, has property (J).✓ Correct
B.Any finite abelian group has property (J).✓ Correct
C.The symmetric group $S_{3}$ has property (J).
D.Any group with 77 elements has property (J).✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Juneshow ▾

Let G be a group of order 2023. Which of the following statements are true?

A.G is an Abelian group.✓ Correct
B.G is a cyclic group.
C.G is a simple group.
D.G is not a simple group.✓ Correct

Solution

$2023 = 7 \cdot 17^{2}. n_{7} \equiv 1 (mod 7)$ and $n_{7}$ | $289 \Rightarrow n_{7} = 1; n_{17}$ | 7 and $\equiv 1 (mod 17) \Rightarrow n_{17} = 1$ .

So $G \cong \mathbb{Z}_{7} \times P$ with P of order $17^{2}$ abelian: G is abelian and not simple.

P may be $\mathbb{Z}_{17} \times \mathbb{Z}_{17}$ , so G need not be cyclic.

Part Cshow ▾

Let G be a group of order 12. Which of the following must be true?

A.At least one Sylow subgroup of G is normal.✓ Correct
B.G has a normal Sylow 2-subgroup.
C.G is solvable.✓ Correct
D.G has a subgroup of order 6.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The number of Sylow 5-subgroups of a group of order 20 can be

A.1 only✓ Correct
B.1 or 4
C.1 or 5
D.4 only

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

Every group of order 15 is

A.cyclic✓ Correct
B.non-abelian
C.isomorphic to a direct product of two non-trivial non-cyclic groups
D.simple

The worked solution and the per-option analysis are in Trap Analysis.

6. Finite abelian groups

Lecture 09 - Types of groups

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Cyclic and abelian groups; the structure theorem itself is in the notes above.

Lecture 24 - Cauchy's theorem

NPTEL · Introduction to Abstract Group Theory (K. Hanumanthu, CMI)

Cauchy's theorem — the partial converse to Lagrange that abelian groups upgrade to a full converse.

Structure of finite abelian groups free

In plain words

Every finite commutative group is a unique stack of cyclic prime-power pieces, so counting them is counting partitions. The fact it rests on is that $C_4 \times C_2$ is not $C_8$ .

The two normal forms

Every finite abelian group is uniquely

invariant factors: $\mathbb{Z}_{d_{1}}$ × ⋯ × $\mathbb{Z}_{d_k}$ with $d_{1}$ | $d_{2}$ | ⋯ | d_k, or
elementary divisors: a product of $\mathbb{Z}_{p^a}$ over prime powers.

The number of abelian groups of order $p^{n}$ is p(n), the number of partitions of n. For a general n, multiply the partition counts of each prime exponent.

Order	# abelian groups
$p^{2}$	2
$p^{3}$	3
$p^{4}$	5
$8 = 2^{3}$	3
16	5
$36 = 2^{2}3^{2}$	2 × 2 = 4
$100 = 2^{2}5^{2}$	4

Counting elements and subgroups

$- \mathbb{Z}_n$ has $\varphi(d)$ elements of order d for each d | n, and exactly one subgroup of order d. $- \mathbb{Z}_p \times \mathbb{Z}_p$ has $p^{2} - 1$ elements of order p and p + 1 subgroups of order p.

An abelian group has a subgroup of every order dividing |G| — the converse of Lagrange holds here.
G is cyclic ⇔ for each d, at most d elements satisfy xᵈ = e ⇔ the invariant factor form has a single factor.

Useful isomorphisms

$\mathbb{Z}_m \times \mathbb{Z}_n \cong$ $\mathbb{Z}_{mn}$ $\Leftrightarrow \gcd(m, n) = 1. (\mathbb{Z}/p\mathbb{Z})$ * is cyclic of order $p - 1. (\mathbb{Z}/2^k)$ * $\cong \mathbb{Z}_{2} \times$ $\mathbb{Z}_{2^{k-2}}$ for k ≥ 3 — so it is not cyclic for k ≥ 3.

Key takeaways

Count abelian groups of order n by multiplying partition numbers of the prime exponents. $- \mathbb{Z}_p \times \mathbb{Z}_p: p + 1$ subgroups of order p, not $p^{2} - 1$ .
Abelian groups satisfy the converse of Lagrange.

Classifying finite abelian groups, and telling two apartinteractive

This visual is part of Trap Analysis

Counting the groups of a given order, and the invariant that distinguishes them in one step.

Exam focus: Convert between invariant-factor and elementary-divisor form fast, and count subgroups/elements of a given order using the partition structure.

Practice

Part C2024 Decembershow ▾

A group G is said to be divisible if for every y ∈ G and for every positive integer n, there exists x ∈ G such that

x^{n} = y

. Which of the following groups are divisible?

A. $\mathbb{Q}$ with ordinary addition✓ Correct
B. $\mathbb{C}$ \ {0} with ordinary multiplication✓ Correct
C.The cyclic group of order 5
D.The symmetric group $S_{5}$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

An abelian group G is said to have property (P) if for any subgroup N of G, there exists a subgroup H of G such that G = N + H and N ∩ H = {0}. Which of the following statements are true?

A.If an abelian group G has property (P), every subgroup of G has property (P).✓ Correct
B.If an abelian group G has property (P), then every element of G has finite order.✓ Correct
C.The group $\mathbb{Z}$ has property (P).
D.If an abelian group G has property (P), then no element has order $p^{2}$ , where p is a prime number.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let G be a finite abelian group of order

360 = 2^{3}\cdot3^{2}\cdot5

. Which of the following are true?

A. $G \cong P_{2} \times P_{3} \times P_{5}, a$ direct product of its Sylow subgroups, with | $P_{2}$ | = 8, | $P_{3}$ | = 9, | $P_{5}$ | = 5.✓ Correct
B.There are exactly 3 isomorphism classes of abelian groups of order $8: \mathbb{Z}_{8}, \mathbb{Z}_{4}\times\mathbb{Z}_{2}, \mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ .✓ Correct
C.The number of abelian groups of order 360, up to isomorphism, is p(3)·p(2)·p(1) = 3·2·1 = 6, where p is the partition function.✓ Correct
D.Every abelian group of order 360 is cyclic.

Solution

(1) abelian Sylow subgroups are automatically normal, so the group splits as their direct product. (2) the number of abelian groups of order $p^{3}$ is the number of partitions of 3. (3) applies this prime-power by prime-power. (4) is false: only one of those 6 isomorphism classes — the one built entirely from cyclic Sylow factors, $\mathbb{Z}_{8}\times\mathbb{Z}_{9}\times\mathbb{Z}_{5} \cong \mathbb{Z}_{360}$ — is actually cyclic; the other 5 are not.

Part Bshow ▾

The number of subgroups of order p in

\mathbb{Z}_p \times \mathbb{Z}_p

A.1
B.p
C.p + 1✓ Correct
D. $p^{2} - 1$

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The number of non-isomorphic abelian groups of order 72 is

A.4
B.6✓ Correct
C.8
D.9

Solution

$72 = 2^{3}\cdot3^{2}: p(3)\cdot{}p(2) = 3\cdot2 = 6$ .

Part Cshow ▾

Let

G = \mathbb{Z}_{4} \times \mathbb{Z}_{6}

. Which of the following are true?

A.G, in invariant factor form, is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{12}$ .✓ Correct
B.The maximum order of an element of G is 12.✓ Correct
C.G is cyclic.
D. $G \cong \mathbb{Z}_{24}$ .

Solution

$\mathbb{Z}_{4}\times\mathbb{Z}_{6} \cong \mathbb{Z}_{4}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3} \cong \mathbb{Z}_{2}\times(\mathbb{Z}_{4}\times\mathbb{Z}_{3}) \cong \mathbb{Z}_{2}\times\mathbb{Z}_{12}$ , since gcd(4,3)=1 lets the Chinese Remainder Theorem combine those two factors — (1). The maximum element order is lcm(4,6) = 12 — (2). (3)–(4) are false: a cyclic group of order 24 needs an element of order 24, but the largest order available here is 12 < 24 — the direct product's factors 4 and 6 are not coprime, so CRT cannot collapse the whole thing to a single cyclic factor.