Permutation Groups and Sylow Theorems Questions and Answers

Q: Let G be a group of prime power order. Then :

The answer is G is solvable.A group of order \(p^{n}\) has a non-trivial centre, so by induction it has a normal series with abelian (indeed cyclic of order \(p\)) factors: every \(p\)-group is nilpotent, hence solvable.The other statements fail in general. The quaternion group of order 8 is a non-abelian 2-group. A \(p\)-group of order \(p^{n}\) with \(n \ge 2\) has normal subgroups of every order \(p^{k}\), so it is not simple. And the Sylow \(p\)-subgroup of a \(p\)-group is the whole group, which need not be abelian (again the quaternion group).

Q: In the group \(S_{3}\), there are exactly :

\(S_{3}\) has 3 elements with \(x^{3} = e\) and 4 elements with \(x^{2} = e\).\(S_{3}\) has six elements: the identity, three transpositions (order 2) and two 3-cycles (order 3).\(x^{3} = e\): the identity and the two 3-cycles, 3 elements.\(x^{2} = e\): the identity and the three transpositions, 4 elements.Options with 2 elements for \(x^{3} = e\), or 3 for \(x^{2} = e\), forget to count the identity, which satisfies every equation \(x^{k} = e\).

Q: Let G be a group of order \(n\), \(p\) a prime such that \(p \mid n\). Then :

The correct statement is that the number of Sylow \(p\)-subgroups equals the index of N(P) in G.The Sylow \(p\)-subgroups are all conjugate (second Sylow theorem). G acts on them by conjugation transitively, and the stabiliser of P is its normaliser N(P). By the orbit-stabiliser theorem, the number of Sylow \(p\)-subgroups is \(|G| / |N(P)| = [G : N(P)]\).The others fail in general. In \(S_{3}\) the Sylow 2-subgroups are not normal (A). A Sylow 2-subgroup of \(S_{4}\) is the dihedral group of order 8, which is not abelian (B). And the index of P itself is usually larger than the number of Sylow subgroups: in \(S_{3}\), \([G : P] = 3\) for \(p = 2\), which happens to match, but for \(p = 3\), \([G : P] = 2\) while there is only one Sylow 3-subgroup (D).

Q: Consider the following statements :(I) A p-sylow subgroup of the underlying additive group of a finite commutative ring R with unity is an ideal of R.(II) If M is a maximal ideal of a commutative ring R and R* be the group of units of R. Such that R is a disjoint union of M and R*, then R has a maximal ideal other than M.Which is true ?

Only (I) is true.(I) The p-Sylow subgroup of the additive group of a finite ring is \(S = \{x : p^{k}x = 0\}\) for a suitable k. It is an additive subgroup, and for any r in R, \(p^{k}(rx) = r(p^{k}x) = 0\), so \(rx \in S\). Hence S is an ideal.(II) If R is the disjoint union of M and the units R*, then every non-unit lies in M. Any proper ideal consists of non-units, so it lies inside M. Hence M is the only maximal ideal (R is local), and the claim that R has another maximal ideal is false.

Q: Consider the following statements :(I) If \(n \ge 5\), then all cycles of length 3 are conjugates in An.(II) Two elements of An that are conjugates in Sn are always conjugates in An also.Which is correct ?

Only (I) is true.(I) Any two 3-cycles are conjugate in \(S_{n}\) by some \(\sigma\). If \(\sigma\) is odd and \(n \ge 5\), there is a transposition \(\tau\) disjoint from the 3-cycle \(c\), and \(\sigma\tau\) is even and conjugates c to the same target, since \(\tau\) commutes with c. So all 3-cycles are conjugate in \(A_{n}\).(II) is false: \(S_{n}\)-classes can split in \(A_{n}\). In \(A_{5}\), the 24 five-cycles form one class in \(S_{5}\) but split into two classes of 12 in \(A_{5}\).

Q: Consider the following statements :(I) Every finite group is isomorphic to a subgroup of the alternating group An for some \(n \gt 1\).(II) The symmetric group Sn is nilpotent for \(n \ge 3\).Which is correct ?

Only (I) is correct.(I) By Cayley's theorem a finite group G embeds in \(S_{m}\) with \(m = |G|\). The map sending a permutation \(\sigma\) of {1, ..., m} to \(\sigma\) itself when \(\sigma\) is even, and to \(\sigma\) times the transposition \((m{+}1\ m{+}2)\) when \(\sigma\) is odd, embeds \(S_{m}\) in \(A_{m+2}\). So G is isomorphic to a subgroup of some \(A_{n}\).(II) \(S_{n}\) is not nilpotent for \(n \ge 3\). A nilpotent group has a non-trivial centre, but the centre of \(S_{n}\) is trivial for \(n \ge 3\); for \(S_{3}\), which has order 6, nilpotency would also make it the product of its Sylow subgroups and hence abelian, which it is not.

Q: The permutation (12) (13) in \(S_{4}\) is similar to :

The permutation is similar (conjugate) to (2 3 4).Compose right to left: (13) sends 1 to 3, and (12) leaves 3 alone, so 1 goes to 3; 3 goes to 1 and then to 2; 2 is fixed by (13) and sent to 1 by (12). So\[ (12)(13) = (1\;3\;2) \]a 3-cycle. In \(S_{n}\) two permutations are conjugate exactly when they have the same cycle type, so (12)(13) is conjugate to every 3-cycle, including (2 3 4). (Composing left to right gives (1 2 3), also a 3-cycle.)(2 3) is a transposition, (1 2 3 4) a 4-cycle and (1 2)(3 4) a product of two disjoint transpositions, all different cycle types.

11.

Let G be a group of prime power order. Then :

A.G is abelian
B.G is simple
C.G is solvable
D.G has a Sylow subgroup which is abelian

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is G is solvable.

A group of order \(p^{n}\) has a non-trivial centre, so by induction it has a normal series with abelian (indeed cyclic of order \(p\)) factors: every \(p\)-group is nilpotent, hence solvable.

The other statements fail in general. The quaternion group of order 8 is a non-abelian 2-group. A \(p\)-group of order \(p^{n}\) with \(n \ge 2\) has normal subgroups of every order \(p^{k}\), so it is not simple. And the Sylow \(p\)-subgroup of a \(p\)-group is the whole group, which need not be abelian (again the quaternion group).

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12.

In the group \(S_{3}\), there are exactly :

A.2 elements satisfying \(x^{3} = e\) and 3 elements satisfying \(x^{2} = e\)
B.3 elements satisfying \(x^{3} = e\) and 4 elements satisfying \(x^{2} = e\)
C.2 elements satisfying \(x^{3} = e\) and 4 elements satisfying \(x^{2} = e\)
D.3 elements satisfying \(x^{3} = e\) and 3 elements satisfying \(x^{2} = e\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

\(S_{3}\) has 3 elements with \(x^{3} = e\) and 4 elements with \(x^{2} = e\).

\(S_{3}\) has six elements: the identity, three transpositions (order 2) and two 3-cycles (order 3).

\(x^{3} = e\): the identity and the two 3-cycles, 3 elements.
\(x^{2} = e\): the identity and the three transpositions, 4 elements.

Options with 2 elements for \(x^{3} = e\), or 3 for \(x^{2} = e\), forget to count the identity, which satisfies every equation \(x^{k} = e\).

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13.

Let G be a group of order \(n\), \(p\) a prime such that \(p \mid n\). Then :

A.every Sylow \(p\)-subgroup of G is normal in G
B.every Sylow \(p\)-subgroup of G is abelian
C.the number of Sylow \(p\)-subgroups of G is equal to the index of N(P) in G, where N(P) is the normaliser of any Sylow \(p\)-subgroup P of G
D.the number of Sylow \(p\)-subgroups of G is equal to the index of P in G, where P is any Sylow \(p\)-subgroup of G

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The correct statement is that the number of Sylow \(p\)-subgroups equals the index of N(P) in G.

The Sylow \(p\)-subgroups are all conjugate (second Sylow theorem). G acts on them by conjugation transitively, and the stabiliser of P is its normaliser N(P). By the orbit-stabiliser theorem, the number of Sylow \(p\)-subgroups is \(|G| / |N(P)| = [G : N(P)]\).

The others fail in general. In \(S_{3}\) the Sylow 2-subgroups are not normal (A). A Sylow 2-subgroup of \(S_{4}\) is the dihedral group of order 8, which is not abelian (B). And the index of P itself is usually larger than the number of Sylow subgroups: in \(S_{3}\), \([G : P] = 3\) for \(p = 2\), which happens to match, but for \(p = 3\), \([G : P] = 2\) while there is only one Sylow 3-subgroup (D).

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14.

Which of the following is true ?

A.Let G be a finite group such that \(|G| = p^{k}\), where \(p\) is prime and \(k \in \mathbf{N}\). Then G has a non-trivial center
B.The group of order \(p^{2}\) (\(p\)-prime) is simple
C.The group of order \(pq\), where \(p\) and \(q\) are primes is simple
D.The group of order 45 is simple

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The true statement is that a finite group of order \(p^{k}\), \(k \ge 1\), has a non-trivial center.

This is the class equation argument. Write \(|G| = |Z(G)| + \sum [G : C(x_{i})]\), the sum over representatives of the non-central conjugacy classes. Each index \([G : C(x_{i})]\) is a power of \(p\) greater than 1, so \(p\) divides the sum, and \(p\) divides \(|G|\). Hence \(p\) divides \(|Z(G)|\), so the center has at least \(p\) elements.

The other statements are false. A group of order \(p^{2}\) is abelian with proper nontrivial subgroups, so it is not simple. A group of order \(pq\) has a normal Sylow subgroup (for \(p \lt q\), the Sylow \(q\)-subgroup is normal), so it is not simple. A group of order 45 = 9 × 5 has exactly one Sylow 5-subgroup (the count divides 9 and is 1 mod 5), which is normal.

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15.

Consider the following statements :

(I) A p-sylow subgroup of the underlying additive group of a finite commutative ring R with unity is an ideal of R.
(II) If M is a maximal ideal of a commutative ring R and R* be the group of units of R. Such that R is a disjoint union of M and R*, then R has a maximal ideal other than M.

Which is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only (I) is true.

(I) The p-Sylow subgroup of the additive group of a finite ring is \(S = \{x : p^{k}x = 0\}\) for a suitable k. It is an additive subgroup, and for any r in R, \(p^{k}(rx) = r(p^{k}x) = 0\), so \(rx \in S\). Hence S is an ideal.

(II) If R is the disjoint union of M and the units R*, then every non-unit lies in M. Any proper ideal consists of non-units, so it lies inside M. Hence M is the only maximal ideal (R is local), and the claim that R has another maximal ideal is false.

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16.

Consider the following statements :

(I) If \(n \ge 5\), then all cycles of length 3 are conjugates in A_n.
(II) Two elements of A_n that are conjugates in S_n are always conjugates in A_n also.

Which is correct ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only (I) is true.

(I) Any two 3-cycles are conjugate in \(S_{n}\) by some \(\sigma\). If \(\sigma\) is odd and \(n \ge 5\), there is a transposition \(\tau\) disjoint from the 3-cycle \(c\), and \(\sigma\tau\) is even and conjugates c to the same target, since \(\tau\) commutes with c. So all 3-cycles are conjugate in \(A_{n}\).

(II) is false: \(S_{n}\)-classes can split in \(A_{n}\). In \(A_{5}\), the 24 five-cycles form one class in \(S_{5}\) but split into two classes of 12 in \(A_{5}\).

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17.

Consider the following statements :

(I) Every finite group is isomorphic to a subgroup of the alternating group A_n for some \(n \gt 1\).
(II) The symmetric group S_n is nilpotent for \(n \ge 3\).

Which is correct ?

A.Only (I) is correct
B.Only (II) is correct
C.Both (I) and (II) are correct
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only (I) is correct.

(I) By Cayley's theorem a finite group G embeds in \(S_{m}\) with \(m = |G|\). The map sending a permutation \(\sigma\) of {1, ..., m} to \(\sigma\) itself when \(\sigma\) is even, and to \(\sigma\) times the transposition \((m{+}1\ m{+}2)\) when \(\sigma\) is odd, embeds \(S_{m}\) in \(A_{m+2}\). So G is isomorphic to a subgroup of some \(A_{n}\).

(II) \(S_{n}\) is not nilpotent for \(n \ge 3\). A nilpotent group has a non-trivial centre, but the centre of \(S_{n}\) is trivial for \(n \ge 3\); for \(S_{3}\), which has order 6, nilpotency would also make it the product of its Sylow subgroups and hence abelian, which it is not.

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18.

The permutation (12) (13) in \(S_{4}\) is similar to :

A.(2 3 4)
B.(2 3)
C.(1 2 3 4)
D.(1 2) (3 4)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The permutation is similar (conjugate) to (2 3 4).

Compose right to left: (13) sends 1 to 3, and (12) leaves 3 alone, so 1 goes to 3; 3 goes to 1 and then to 2; 2 is fixed by (13) and sent to 1 by (12). So

\[ (12)(13) = (1\;3\;2) \]

a 3-cycle. In \(S_{n}\) two permutations are conjugate exactly when they have the same cycle type, so (12)(13) is conjugate to every 3-cycle, including (2 3 4). (Composing left to right gives (1 2 3), also a 3-cycle.)

(2 3) is a transposition, (1 2 3 4) a 4-cycle and (1 2)(3 4) a product of two disjoint transpositions, all different cycle types.

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