Which of the following groups are cyclic?

U(9), the units of ℤ₉ under multiplication

U(8), the units of ℤ₈ under multiplication

Which statements are true for finite groups?

every group of order p², p prime, is abelian

if G/Z(G) is cyclic then G is abelian

every subgroup of index 2 is normal

every group of order p², p prime, is abelian

if G/Z(G) is cyclic then G is abelian

every subgroup of index 2 is normal

Let G be a group of order 15. Which statements are true?

G has a normal subgroup of order 5

G has exactly 8 elements of order 15

G has a normal subgroup of order 5

G has exactly 8 elements of order 15

Let G be a group of order 12 with exactly four Sylow 3-subgroups. Then

G has exactly 8 elements of order 3 and a unique, hence normal, Sylow 2-subgroup

G has a normal Sylow 3-subgroup

G has exactly three Sylow 2-subgroups

Groups: Homomorphisms, Permutations, Group Actions and the Sylow Theorems

Algebra is Section 6 of the GATE Mathematics (MA) paper, and its group half is the richest source of counting questions on the whole paper: the number of generators of a cyclic group, the number of elements of a given order, the number of homomorphisms between two cyclic groups, the number of abelian groups of a given order, the largest order of a permutation, the number of Sylow subgroups. Every one of these has a formula, and every formula rests on one of five results — Lagrange’s theorem, the structure of cyclic groups, the first isomorphism theorem, the orbit–stabiliser theorem and the Sylow theorems. This chapter states each with its hypotheses, gives the formula it produces, and pairs it with the counterexample the paper uses to test the converse: A₄, of order 12, which has no subgroup of order 6. The second chapter takes rings, ideals, factorisation domains and fields.

1. Subgroups, cosets, Lagrange and cyclic groups

A non-empty H ⊆ G is a subgroup iff ab⁻¹ ∈ H for all a, b ∈ H (for finite H, closure suffices). The left cosets of H partition G into pieces of size |H|, so Lagrange: |H| divides |G|, and the order of every element divides |G|; hence a group of prime order is cyclic, and a^|G| = e. The converse fails: A₄ has order 12 but no subgroup of order 6.

Counting in a cyclic group ℤₙ
Quantity	Formula	Example
generators	φ(n)	ℤ₂₀: φ(20) = 8
subgroups	exactly one of each order d \| n, namely ⟨n/d⟩	ℤ₁₂ has 6 subgroups
elements of order d	φ(d) if d \| n, else 0	ℤ₃₀ has φ(6) = 2 elements of order 6
order of a	n/gcd(a, n)	8 in ℤ₁₂ has order 3
homomorphisms ℤₘ → ℤₙ	gcd(m, n)	ℤ₁₂ → ℤ₁₈: 6
automorphisms of ℤₙ	Aut(ℤₙ) ≅ U(n), of order φ(n)	\|Aut(ℤ₁₅)\| = 8

The group of units U(n) = {a ∈ ℤₙ : gcd(a, n) = 1} under multiplication has order φ(n), and it is cyclic exactly when n = 1, 2, 4, pᵏ or 2pᵏ for an odd prime p. So U(9) ≅ ℤ₆ is cyclic, while U(8) = {1, 3, 5, 7} has every element of order ≤ 2 and is ℤ₂ × ℤ₂. ℤₘ × ℤₙ is cyclic iff gcd(m, n) = 1.

2. Normal subgroups, quotients, homomorphisms and automorphisms

N is normal (N ⊴ G) if gNg⁻¹ = N for all g; then the cosets form the quotient group G/N of order [G : N]. The kernel of a homomorphism is normal, and the first isomorphism theorem gives G/ker φ ≅ φ(G); so |φ(G)| divides both |G| and the order of the target. A homomorphism ℤₘ → H is fixed by the image of 1, which may be any element of H whose order divides m — which is why there are gcd(m, n) homomorphisms ℤₘ → ℤₙ.

Every subgroup of index 2 is normal; every subgroup of an abelian group is normal; the centre Z(G) is normal.
If G/Z(G) is cyclic then G is abelian, so G/Z(G) is never cyclic of order > 1; and Inn(G) ≅ G/Z(G).
Normality is not transitive: in D₄ (order 8), a subgroup of order 2 generated by a reflection is normal in a Klein four-subgroup, which is normal in D₄, but it is not normal in D₄.

3. Permutation groups

Every permutation is a product of disjoint cycles, uniquely up to order; its order is the lcm of the cycle lengths and its sign is (−1)^{(sum of (length − 1))}: a k-cycle is even exactly when k is odd. Aₙ, the even permutations, has index 2 in Sₙ. Two permutations are conjugate in Sₙ iff they have the same cycle type, so conjugacy classes of Sₙ correspond to partitions of n; the 3-cycles of S₅ form one class of size (5·4·3)/3 = 20. Cayley: every group of order n embeds in Sₙ.

Largest order of an element of Sₙ: maximise the lcm over partitions of n
n	Best cycle type	Maximum order
5	(2, 3)	6
6	(1, 2, 3) or (6)	6
7	(3, 4)	12
8	(3, 5)	15

4. Group actions and finite abelian groups

An action of G on a set X gives orbits partitioning X and stabilisers Gₓ ≤ G, with the orbit–stabiliser theorem |G·x| = [G : Gₓ]. For the conjugation action this is the class equation |G| = |Z(G)| + Σ [G : C_G(xᵢ)], the sum over representatives of the non-central classes. Consequences: a non-trivial p-group has a non-trivial centre, and every group of order p² is abelian (≅ ℤ_p² or ℤₚ × ℤₚ).

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. So the number of abelian groups of order p₁^e₁⋯p_k^e_{k} is P(e₁)⋯P(e_k), P the partition function (P(1) = 1, P(2) = 2, P(3) = 3, P(4) = 5, P(5) = 7). Order 72 = 2³·3² gives 3 × 2 = 6. Elements of order dividing 2 in ℤ₂ × ℤ₄ × ℤ₆ number 2 × 2 × 2 = 8, so 7 of them have order exactly 2.

5. The Sylow theorems and their applications

Let |G| = pᵃm with p prime and p ∤ m. (I) G has a subgroup of order pᵃ (a Sylow p-subgroup), and indeed of every order pᵇ, b ≤ a. (II) Any two Sylow p-subgroups are conjugate, and every p-subgroup lies in one. (III) The number n_p of Sylow p-subgroups satisfies n_p ≡ 1 (mod p) and n_p | m. A Sylow p-subgroup is normal iff n_p = 1.

Standard applications
Order	Sylow count	Conclusion
15 = 3·5	n₃ \| 5, n₃ ≡ 1 (3) ⇒ 1; n₅ \| 3, n₅ ≡ 1 (5) ⇒ 1	both normal, G ≅ ℤ₃ × ℤ₅ ≅ ℤ₁₅; in general pq with p < q and p ∤ q − 1 is cyclic
12	n₃ ∈ {1, 4}; if 4, they give 8 elements of order 3	the remaining 4 elements form the unique Sylow 2-subgroup, so some Sylow subgroup is normal (A₄ has n₃ = 4)
S₅, order 120	24 five-cycles, 4 in each Sylow 5-subgroup	n₅ = 6
168, G simple	n₇ \| 24, n₇ ≡ 1 (7), n₇ ≠ 1 ⇒ n₇ = 8	8 × 6 = 48 elements of order 7

🧠 Counting elements to force a normal subgroup

Distinct subgroups of prime order p meet only in e, so n_p Sylow p-subgroups of order p contribute n_p(p − 1) elements of order p. If these counts over two primes exceed |G|, one n_p must be 1. For |G| = 30: n₃ = 10 and n₅ = 6 would give 20 + 24 = 44 > 30 elements, so a group of order 30 is never simple.

Key takeaways

Lagrange: subgroup orders and element orders divide |G|; the converse fails for A₄ and 6.
In ℤₙ: φ(n) generators, one subgroup per divisor, φ(d) elements of order d; gcd(m, n) homomorphisms ℤₘ → ℤₙ; |Aut ℤₙ| = φ(n).
U(n) is cyclic exactly for n = 1, 2, 4, pᵏ, 2pᵏ; ℤₘ × ℤₙ is cyclic iff gcd(m, n) = 1.
A permutation’s order is the lcm of its cycle lengths; conjugacy in Sₙ is cycle type; abelian groups of order n are counted by partitions of the prime exponents.
Sylow: n_p ≡ 1 (mod p) and n_p divides the p′-part; n_p = 1 means normal; count elements of prime order to force it.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The number of generators of the cyclic group ℤ₂₀ is ____.
Numerical answer — type the value.
Show answer
Answer: 8
k generates ℤ₂₀ iff gcd(k, 20) = 1, so the count is φ(20) = 20 · (1 − 1/2)(1 − 1/5) = 8: namely 1, 3, 7, 9, 11, 13, 17, 19.
The number of elements of order 2 in the group ℤ₂ × ℤ₄ × ℤ₆ is ____.
Numerical answer — type the value.
Show answer
Answer: 7
(a, b, c) has order dividing 2 iff 2a = 0, 2b = 0, 2c = 0, which has 2 solutions in each factor (0 and the element of order 2), so 2³ = 8 elements. Removing the identity leaves 7 of order exactly 2.
The number of group homomorphisms from ℤ₁₂ to ℤ₁₈ is ____.
Numerical answer — type the value.
Show answer
Answer: 6
A homomorphism is determined by x = φ(1), which must satisfy 12x = 0 in ℤ₁₈, i.e. 18 | 12x, i.e. 3 | 2x, i.e. x ∈ {0, 3, 6, 9, 12, 15}. That is gcd(12, 18) = 6 homomorphisms. None is injective, because an image has order dividing both 12 and 18, hence at most 6.
Which of the following groups are cyclic?
1. U(9), the units of ℤ₉ under multiplication
2. U(8), the units of ℤ₈ under multiplication
3. ℤ₄ × ℤ₆
4. ℤ₃ × ℤ₁₀
Show answer
Answer: A — U(9), the units of ℤ₉ under multiplication; D — ℤ₃ × ℤ₁₀
(1) 9 = 3², so U(9) is cyclic of order 6, generated by 2 (2, 4, 8, 7, 5, 1). (2) U(8) = {1, 3, 5, 7} with every square 1: ℤ₂ × ℤ₂. (3) gcd(4, 6) = 2, so the largest element order is lcm = 12 < 24. (4) gcd(3, 10) = 1, so ℤ₃ × ℤ₁₀ ≅ ℤ₃₀.
The order of the automorphism group of the cyclic group ℤ₁₅ is ____.
Numerical answer — type the value.
Show answer
Answer: 8
An automorphism sends the generator 1 to another generator, and every generator gives one, so Aut(ℤ₁₅) ≅ U(15) with order φ(15) = φ(3)φ(5) = 2 × 4 = 8.
The largest order of an element of the symmetric group S₇ is ____.
Numerical answer — type the value.
Show answer
Answer: 12
The order of a permutation is the lcm of its cycle lengths, which partition 7. The candidates: 7 → 7; 6 + 1 → 6; 5 + 2 → 10; 4 + 3 → 12; 4 + 2 + 1 → 4; 3 + 2 + 2 → 6. The maximum is 12, from a 4-cycle times a disjoint 3-cycle.
Which of the following permutations in S₆ is even?
1. (1 2 3 4)(5 6)
2. (1 2 3 4)
3. (1 2)(3 4)(5 6)
4. (1 2 3 4 5 6)
Show answer
Answer: A — (1 2 3 4)(5 6)
A k-cycle is a product of k − 1 transpositions. (1 2 3 4)(5 6) uses 3 + 1 = 4, even. (1 2 3 4) uses 3, odd; (1 2)(3 4)(5 6) uses 3, odd; a 6-cycle uses 5, odd.
Which statements are true for finite groups?
1. every group of order p², p prime, is abelian
2. if G/Z(G) is cyclic then G is abelian
3. every subgroup of index 2 is normal
4. A₄ has a subgroup of order 6
Show answer
Answer: A — every group of order p², p prime, is abelian; B — if G/Z(G) is cyclic then G is abelian; C — every subgroup of index 2 is normal
(2) If G/Z = ⟨gZ⟩, every element is gⁱz, and such elements commute. (1) By the class equation Z(G) ≠ {e}, so |G/Z| is 1 or p, cyclic, and (2) applies. (3) The two left cosets are H and G ∖ H, and so are the two right cosets. (4) is false: A₄ has 8 elements of order 3, and a subgroup of order 6 (index 2, normal) would contain all their squares, hence all 8.
The number of abelian groups of order 72, up to isomorphism, is ____.
Numerical answer — type the value.
Show answer
Answer: 6
72 = 2³ · 3². The 2-part is ℤ₈, ℤ₄ × ℤ₂ or ℤ₂³ (partitions of 3: three), and the 3-part is ℤ₉ or ℤ₃² (partitions of 2: two). The fundamental theorem gives 3 × 2 = 6.
The number of elements in the conjugacy class of the 3-cycle (1 2 3) in S₅ is ____.
Numerical answer — type the value.
Show answer
Answer: 20
Conjugacy in S₅ is cycle type, so the class is all 3-cycles: choose 3 of 5 symbols (10 ways) and arrange them cyclically (2 ways), 20 in all. By orbit–stabiliser the centraliser has order 120/20 = 6, namely ⟨(1 2 3)⟩ × ⟨(4 5)⟩.
The number of Sylow 5-subgroups of the symmetric group S₅ is ____.
Numerical answer — type the value.
Show answer
Answer: 6
|S₅| = 120 = 2³ · 3 · 5, so a Sylow 5-subgroup has order 5 and is generated by a 5-cycle. There are 4!/1 = 24 five-cycles and each subgroup contains 4 of them, so n₅ = 24/4 = 6, consistent with n₅ ≡ 1 (mod 5) and n₅ | 24.
Let G be a group of order 15. Which statements are true?
1. G is cyclic
2. G has a normal subgroup of order 5
3. G can be non-abelian
4. G has exactly 8 elements of order 15
Show answer
Answer: A — G is cyclic; B — G has a normal subgroup of order 5; D — G has exactly 8 elements of order 15
n₅ divides 3 and is ≡ 1 (mod 5), so n₅ = 1; likewise n₃ divides 5 and is ≡ 1 (mod 3), so n₃ = 1. Both Sylow subgroups are normal with trivial intersection, so G ≅ ℤ₃ × ℤ₅ ≅ ℤ₁₅: cyclic, hence abelian, with φ(15) = 8 generators, the elements of order 15.
Let G be a group of order 12 with exactly four Sylow 3-subgroups. Then
1. G has exactly 8 elements of order 3 and a unique, hence normal, Sylow 2-subgroup
2. G has a normal Sylow 3-subgroup
3. G is abelian
4. G has exactly three Sylow 2-subgroups
Show answer
Answer: A — G has exactly 8 elements of order 3 and a unique, hence normal, Sylow 2-subgroup
Four subgroups of order 3 meeting pairwise in e give 4 × 2 = 8 elements of order 3. The other 4 elements must form any Sylow 2-subgroup (order 4), so there is exactly one and it is normal. n₃ = 4 ≠ 1 means no normal Sylow 3-subgroup, so G is not abelian; A₄ is the example.
Let G be a simple group of order 168. The number of elements of order 7 in G is ____.
Numerical answer — type the value.
Show answer
Answer: 48
168 = 2³ · 3 · 7. n₇ divides 24 and n₇ ≡ 1 (mod 7), so n₇ ∈ {1, 8}; simplicity rules out 1, so n₇ = 8. Subgroups of prime order 7 meet only in e, so they contribute 8 × 6 = 48 elements of order 7.