Permutation Groups and Sylow Theorems Questions and Answers

Q: The number of 5-Sylow subgroup(s) in a group of order 45 is

A group of order 45 has exactly 1 Sylow 5-subgroup.Write \(45 = 3^{2}\cdot 5\). By Sylow's theorems the number \(n_{5}\) of Sylow 5-subgroups divides 9 and satisfies \(n_{5} \equiv 1 \pmod 5\).The divisors of 9 are 1, 3 and 9. Of these, 3 and 9 leave remainders 3 and 4 on division by 5, so \(n_{5} = 1\), and the Sylow 5-subgroup is normal.2 and 4 do not divide 9, and 3 is not congruent to 1 mod 5, so all three other options are impossible. (The same argument gives \(n_{3} = 1\), so every group of order 45 is abelian.)

Q: In the permutation group \(S_{6}\), the number of elements of order 8 is

There are 0 elements of order 8.The order of a permutation is the least common multiple of the lengths of the cycles in its disjoint cycle decomposition. In \(S_{6}\) the cycle lengths add up to at most 6.For the order to be 8, some cycle length would have to be divisible by 8 (an lcm of numbers each at most 6 is divisible by 8 only if one of them is), which needs a cycle of length at least 8. That is impossible with 6 points. The largest power of 2 that occurs as an order in \(S_{6}\) is 4, from a 4-cycle.So the counts 1, 2 and 4 are all wrong; the smallest symmetric group with an element of order 8 is \(S_{8}\).

Q: Which one of the following statements is true?

The true statement is Every group of order 12 has a non-trivial proper normal subgroup.Let \(G\) have order \(12 = 2^{2} \cdot 3\). By Sylow's theorems the number of Sylow 3-subgroups is 1 or 4. If it is 1, that subgroup is normal. If it is 4, they meet only in the identity and supply \(4 \times 2 = 8\) elements of order 3, leaving exactly 4 other elements, which must form the only Sylow 2-subgroup, and that is then normal. Either way \(G\) is not simple.That makes 'some group of order 12 has no such subgroup' false. The alternating group \(A_{4}\) has order 12 and no subgroup of order 6, so the third statement fails, and only the cyclic group of order 12 has an element of order 12, so the fourth fails.

Q: Given a prime number \(p\), let \(n_{p}(G)\) denote the number of \(p\)-Sylow subgroups of a finite group \(G\). Which one of the following is TRUE for every group \(G\) of order 2024?

For every group of order 2024, \(n_{11}(G) \in \{1, 23\}\) and \(n_{23}(G) = 1\).Factor \(2024 = 2^{3}\cdot 11\cdot 23\). By Sylow's theorems:\(n_{23}\) divides 88 and \(n_{23} \equiv 1 \pmod{23}\). The divisors of 88 are 1, 2, 4, 8, 11, 22, 44, 88, and only 1 is congruent to 1 mod 23 (88 leaves remainder 19). So \(n_{23} = 1\).\(n_{11}\) divides 184 and \(n_{11} \equiv 1 \pmod{11}\). The divisors of 184 are 1, 2, 4, 8, 23, 46, 92, 184, and those congruent to 1 mod 11 are 1 and 23. So \(n_{11} \in \{1, 23\}\).The options with \(n_{23} = 11\) or \(n_{23} \in \{1, 88\}\) contradict the first count, and \(n_{11}\) cannot be fixed at a single value for every group: the cyclic group has \(n_{11} = 1\).

Q: Let G be a group of order 595. Which one of the following is TRUE?

The true statement is G must have a proper normal subgroup.Factor the order: \(595 = 5 \times 7 \times 17\). By Sylow's theorems the number of Sylow 17-subgroups divides \(35\) and is congruent to 1 mod 17. The divisors of 35 are 1, 5, 7 and 35, and only 1 is congruent to 1 mod 17, so the Sylow 17-subgroup is unique and therefore normal. It is proper, so G is not simple.The other options fail. Cauchy's theorem gives an element of order 17 because 17 divides 595. The number of Sylow 5-subgroups divides 119 and is 1 mod 5; of 1, 7, 17 and 119 only 1 qualifies, so it is 1, not 17.

Q: Let G be a non-abelian group with n elements, then which of the following values of n is possible :

The possible value is \(n = 10\).The dihedral group \(D_{5}\), the symmetries of a regular pentagon, has 10 elements and is non-abelian: a rotation and a reflection do not commute.Each of the other orders forces every group to be abelian. A group of prime order 13 is cyclic. A group of order \(9 = 3^{2}\) has order \(p^{2}\), and every group of order \(p^{2}\) is abelian. A group of order \(15 = 3 \times 5\) is cyclic, because 3 does not divide \(5 - 1 = 4\), so the Sylow theorems leave only one Sylow subgroup of each order, both normal.

Q: For which of the following values of \(n\) is every group of order \(n\) abelian ?

Every group of order 121 is abelian.\(121 = 11^{2}\), and every group of order \(p^{2}\) is abelian: its centre is non-trivial (a \(p\)-group has non-trivial centre), and if \(G/Z(G)\) were cyclic G would be abelian, so \(Z(G) = G\). Such a group is \(\mathbf{Z}_{121}\) or \(\mathbf{Z}_{11} \times \mathbf{Z}_{11}\).Each other order has a non-abelian group: 100 has \(D_{50}\), the dihedral group of order 100; 24 has \(S_{4}\); and 120 has \(S_{5}\).

Q: There is a non-abelian group of order :

There is a non-abelian group of order 12.The alternating group \(A_{4}\) and the dihedral group \(D_{6}\) of the hexagon both have order 12 and are non-abelian.The other orders force commutativity. 41 is prime, so the group is cyclic. 49 = \(7^{2}\), and every group of order \(p^{2}\) is abelian. 15 = 3 x 5 with 3 not dividing \(5 - 1\), so the only group of order 15 is cyclic.

Q: Every finite group is isomorphic to a subgroup of :

Every finite group is isomorphic to a subgroup of \(S_{n}\) for some \(n\).This is Cayley's theorem: a group G acts on itself by left multiplication, \(g \mapsto (x \mapsto gx)\), and this gives an injective homomorphism from G into the symmetric group on the \(|G|\) elements of G. So G is isomorphic to a subgroup of \(S_{n}\) with \(n = |G|\).The other options are abelian groups, and their subgroups are abelian, so a non-abelian group such as \(S_{3}\) cannot embed in a cyclic group, in \(\mathbf{Z}_{n}\) or in \((\mathbf{R}, +)\).

Q: Let \(\sigma\) be a permutation in \(S_{n}\). Let \(\sigma = \sigma_{1}, \sigma_{2} \cdots \sigma_{k}\), where \(\sigma_{1}, \sigma_{2}, \ldots, \sigma_{k}\) are disjoint cycles of lengths \(l_{1}, l_{2}, \ldots, l_{k}\) respectively. Then the order of \(\sigma\) is equal to :

The order is \(lcm(l_{1}, \ldots, l_{k})\).Disjoint cycles commute, so \(\sigma^{m} = \sigma_{1}^{m}\sigma_{2}^{m}\cdots\sigma_{k}^{m}\), and since they move different points, \(\sigma^{m}\) is the identity only if each \(\sigma_{i}^{m}\) is. A cycle of length \(l_{i}\) has order \(l_{i}\), so \(m\) must be a multiple of every \(l_{i}\); the least such \(m\) is the lcm.For example, \((1\,2)(3\,4\,5)\) has order 6, which is not the sum 5 or the gcd 1; and \((1\,2)(3\,4)\) has order 2, not the product 4.

The number of 5-Sylow subgroup(s) in a group of order 45 is

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

A group of order 45 has exactly 1 Sylow 5-subgroup.

Write \(45 = 3^{2}\cdot 5\). By Sylow's theorems the number \(n_{5}\) of Sylow 5-subgroups divides 9 and satisfies \(n_{5} \equiv 1 \pmod 5\).

The divisors of 9 are 1, 3 and 9. Of these, 3 and 9 leave remainders 3 and 4 on division by 5, so \(n_{5} = 1\), and the Sylow 5-subgroup is normal.

2 and 4 do not divide 9, and 3 is not congruent to 1 mod 5, so all three other options are impossible. (The same argument gives \(n_{3} = 1\), so every group of order 45 is abelian.)

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In the permutation group \(S_{6}\), the number of elements of order 8 is

Exam Year(s)

GATE Mathematics - 2018,

Answer: A

There are 0 elements of order 8.

The order of a permutation is the least common multiple of the lengths of the cycles in its disjoint cycle decomposition. In \(S_{6}\) the cycle lengths add up to at most 6.

For the order to be 8, some cycle length would have to be divisible by 8 (an lcm of numbers each at most 6 is divisible by 8 only if one of them is), which needs a cycle of length at least 8. That is impossible with 6 points. The largest power of 2 that occurs as an order in \(S_{6}\) is 4, from a 4-cycle.

So the counts 1, 2 and 4 are all wrong; the smallest symmetric group with an element of order 8 is \(S_{8}\).

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Which one of the following statements is true?

A.Every group of order 12 has a non-trivial proper normal subgroup
B.Some group of order 12 does not have a non-trivial proper normal subgroup
C.Every group of order 12 has a subgroup of order 6
D.Every group of order 12 has an element of order 12

Exam Year(s)

GATE Mathematics - 2018,

Answer: A

The true statement is Every group of order 12 has a non-trivial proper normal subgroup.

Let \(G\) have order \(12 = 2^{2} \cdot 3\). By Sylow's theorems the number of Sylow 3-subgroups is 1 or 4. If it is 1, that subgroup is normal. If it is 4, they meet only in the identity and supply \(4 \times 2 = 8\) elements of order 3, leaving exactly 4 other elements, which must form the only Sylow 2-subgroup, and that is then normal. Either way \(G\) is not simple.

That makes 'some group of order 12 has no such subgroup' false. The alternating group \(A_{4}\) has order 12 and no subgroup of order 6, so the third statement fails, and only the cyclic group of order 12 has an element of order 12, so the fourth fails.

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Given a prime number \(p\), let \(n_{p}(G)\) denote the number of \(p\)-Sylow subgroups of a finite group \(G\). Which one of the following is TRUE for every group \(G\) of order 2024?

A.\(n_{11}(G) = 1\) and \(n_{23}(G) = 11\)
B.\(n_{11}(G) \in \{1, 23\}\) and \(n_{23}(G) = 1\)
C.\(n_{11}(G) = 23\) and \(n_{23}(G) \in \{1, 88\}\)
D.\(n_{11}(G) = 23\) and \(n_{23}(G) = 11\)

Exam Year(s)

GATE Mathematics - 2024,

Answer: B

For every group of order 2024, \(n_{11}(G) \in \{1, 23\}\) and \(n_{23}(G) = 1\).

Factor \(2024 = 2^{3}\cdot 11\cdot 23\). By Sylow's theorems:

\(n_{23}\) divides 88 and \(n_{23} \equiv 1 \pmod{23}\). The divisors of 88 are 1, 2, 4, 8, 11, 22, 44, 88, and only 1 is congruent to 1 mod 23 (88 leaves remainder 19). So \(n_{23} = 1\).

\(n_{11}\) divides 184 and \(n_{11} \equiv 1 \pmod{11}\). The divisors of 184 are 1, 2, 4, 8, 23, 46, 92, 184, and those congruent to 1 mod 11 are 1 and 23. So \(n_{11} \in \{1, 23\}\).

The options with \(n_{23} = 11\) or \(n_{23} \in \{1, 88\}\) contradict the first count, and \(n_{11}\) cannot be fixed at a single value for every group: the cyclic group has \(n_{11} = 1\).

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Let G be a group of order 595. Which one of the following is TRUE?

A.G cannot have a proper normal subgroup.
B.G must have a proper normal subgroup.
C.G cannot have an element of order 17.
D.The number of Sylow 5-subgroups is 17.

Exam Year(s)

GATE Mathematics - 2026,

Answer: B

The true statement is G must have a proper normal subgroup.

Factor the order: \(595 = 5 \times 7 \times 17\). By Sylow's theorems the number of Sylow 17-subgroups divides \(35\) and is congruent to 1 mod 17. The divisors of 35 are 1, 5, 7 and 35, and only 1 is congruent to 1 mod 17, so the Sylow 17-subgroup is unique and therefore normal. It is proper, so G is not simple.

The other options fail. Cauchy's theorem gives an element of order 17 because 17 divides 595. The number of Sylow 5-subgroups divides 119 and is 1 mod 5; of 1, 7, 17 and 119 only 1 qualifies, so it is 1, not 17.

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Let G be a non-abelian group with n elements, then which of the following values of n is possible :

A.\(n = 13\)
B.\(n = 9\)
C.\(n = 10\)
D.\(n = 15\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The possible value is \(n = 10\).

The dihedral group \(D_{5}\), the symmetries of a regular pentagon, has 10 elements and is non-abelian: a rotation and a reflection do not commute.

Each of the other orders forces every group to be abelian. A group of prime order 13 is cyclic. A group of order \(9 = 3^{2}\) has order \(p^{2}\), and every group of order \(p^{2}\) is abelian. A group of order \(15 = 3 \times 5\) is cyclic, because 3 does not divide \(5 - 1 = 4\), so the Sylow theorems leave only one Sylow subgroup of each order, both normal.

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For which of the following values of \(n\) is every group of order \(n\) abelian ?

A.\(n = 100\)
B.\(n = 121\)
C.\(n = 24\)
D.\(n = 120\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

Every group of order 121 is abelian.

\(121 = 11^{2}\), and every group of order \(p^{2}\) is abelian: its centre is non-trivial (a \(p\)-group has non-trivial centre), and if \(G/Z(G)\) were cyclic G would be abelian, so \(Z(G) = G\). Such a group is \(\mathbf{Z}_{121}\) or \(\mathbf{Z}_{11} \times \mathbf{Z}_{11}\).

Each other order has a non-abelian group: 100 has \(D_{50}\), the dihedral group of order 100; 24 has \(S_{4}\); and 120 has \(S_{5}\).

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There is a non-abelian group of order :

A.49
B.41
C.15
D.12

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

There is a non-abelian group of order 12.

The alternating group \(A_{4}\) and the dihedral group \(D_{6}\) of the hexagon both have order 12 and are non-abelian.

The other orders force commutativity. 41 is prime, so the group is cyclic. 49 = \(7^{2}\), and every group of order \(p^{2}\) is abelian. 15 = 3 x 5 with 3 not dividing \(5 - 1\), so the only group of order 15 is cyclic.

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Every finite group is isomorphic to a subgroup of :

A.a cyclic group
B.\(\mathbf{Z}_{n}\) for some \(n \ge 1\)
C.\(S_{n}\) for some \(n \ge 1\)
D.\((\mathbf{R}, +)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

Every finite group is isomorphic to a subgroup of \(S_{n}\) for some \(n\).

This is Cayley's theorem: a group G acts on itself by left multiplication, \(g \mapsto (x \mapsto gx)\), and this gives an injective homomorphism from G into the symmetric group on the \(|G|\) elements of G. So G is isomorphic to a subgroup of \(S_{n}\) with \(n = |G|\).

The other options are abelian groups, and their subgroups are abelian, so a non-abelian group such as \(S_{3}\) cannot embed in a cyclic group, in \(\mathbf{Z}_{n}\) or in \((\mathbf{R}, +)\).

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10.

Let \(\sigma\) be a permutation in \(S_{n}\). Let \(\sigma = \sigma_{1}, \sigma_{2} \cdots \sigma_{k}\), where \(\sigma_{1}, \sigma_{2}, \ldots, \sigma_{k}\) are disjoint cycles of lengths \(l_{1}, l_{2}, \ldots, l_{k}\) respectively. Then the order of \(\sigma\) is equal to :

A.\(l_{1} + l_{2} + \cdots + l_{k}\)
B.\(l_{1} . l_{2} \cdots l_{k}\)
C.\(gcd(l_{1}, \ldots, l_{k})\)
D.\(lcm(l_{1}, \ldots, l_{k})\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The order is \(lcm(l_{1}, \ldots, l_{k})\).

Disjoint cycles commute, so \(\sigma^{m} = \sigma_{1}^{m}\sigma_{2}^{m}\cdots\sigma_{k}^{m}\), and since they move different points, \(\sigma^{m}\) is the identity only if each \(\sigma_{i}^{m}\) is. A cycle of length \(l_{i}\) has order \(l_{i}\), so \(m\) must be a multiple of every \(l_{i}\); the least such \(m\) is the lcm.

For example, \((1\,2)(3\,4\,5)\) has order 6, which is not the sum 5 or the gcd 1; and \((1\,2)(3\,4)\) has order 2, not the product 4.

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