Groups and Subgroups Questions and Answers

Q: Which of the following groups is not solvable ?

The group that is not solvable is \(S_{5}\).\(S_{5}\) contains the alternating group \(A_{5}\), which is simple and non-abelian. Its derived series is \(S_{5} \supset A_{5} \supset A_{5} \supset \cdots\), never reaching the identity, so \(S_{5}\) is not solvable. This is the group-theoretic reason the general quintic is not solvable by radicals.The others are solvable. \(\mathbf{Z}_{150}\) is abelian. \(S_{4} \supset A_{4} \supset V_{4} \supset \{e\}\) has abelian factors. Every group of order 27 is a \(p\)-group, and every finite \(p\)-group is nilpotent, hence solvable, abelian or not.

Q: Let H be a proper subgroup of the additive group (Q, +) of rational numbers. Then the quotient group Q/H must be :

Q/H must be infinite.(Q, +) is divisible: for every \(q\) and every positive integer \(n\) there is \(r\) with \(nr = q\). A quotient of a divisible group is divisible, so Q/H is divisible. It is also non-trivial because H is proper.A non-trivial finite group of order \(m\) cannot be divisible: if \(g\) is not the identity, there is no \(x\) with \(mx = g\), since \(mx = 0\) for every element. So Q/H cannot be finite, of either parity.For example, with H = Z the quotient Q/Z is infinite: the classes of \(1/n\) for \(n = 1, 2, 3, \ldots\) are all different.

Q: Let G be a group of order 61. Then the number of subgroups of G is :

G has 2 subgroups.61 is prime, so by Lagrange's theorem a subgroup has order 1 or 61. The only subgroup of order 1 is {e}, and the only subgroup of order 61 is G itself. (A group of prime order is cyclic, generated by any non-identity element.)61 counts the elements, not the subgroups; 1 forgets one of the trivial subgroups; and 7 has no connection with 61.

Q: Let G be the group of invertible \(2 \times 2\) matrices over the field \(\mathbf{Z}_{2} = \{0, 1\}\). Then the number of elements in G is :

The group GL(2, \(\mathbf{Z}_{2}\)) has 6 elements.Choose the first column as any non-zero vector in \(\mathbf{Z}_{2}^{2}\): \(4 - 1 = 3\) ways. The second column must not be a multiple of the first, which rules out 0 and the first column itself: \(4 - 2 = 2\) ways. So\[ \begin{aligned} |GL(2, \mathbf{Z}_{2})| &= (2^{2} - 1)(2^{2} - 2) \\ &= 3 \cdot 2 = 6 \end{aligned} \]There are \(2^{4} = 16\) matrices in all, and 10 of them are singular, leaving 6. The options 2, 4 and 8 undercount; the group is in fact isomorphic to \(S_{3}\), acting on the three non-zero vectors.

Q: The group \(\mathbf{Z}_{72}\) is the direct product of groups as :

\(\mathbf{Z}_{72} \cong\) \(\mathbf{Z}_{9} \times \mathbf{Z}_{8}\).By the Chinese remainder theorem, \(\mathbf{Z}_{mn} \cong \mathbf{Z}_{m} \times \mathbf{Z}_{n}\) exactly when \(\gcd(m, n) = 1\). Here \(72 = 9 \times 8\) with \(\gcd(9, 8) = 1\).The other products are not cyclic, because their factors share a common divisor: \(\mathbf{Z}_{2}^{3} \times \mathbf{Z}_{3}^{2}\) has exponent 6; \(\mathbf{Z}_{36} \times \mathbf{Z}_{36}\) has order 1296, not 72; and \(\mathbf{Z}_{2} \times \mathbf{Z}_{36}\) has order 72 but no element of order 72 (its exponent is 36).

Q: Let \(\phi : G \to G'\) be a surjective group homomorphism. Then :

The correct statement is if G is cyclic, then G' is cyclic.If \(G = \langle g \rangle\) and \(\phi\) is onto, every element of G' is \(\phi(g^{k}) = \phi(g)^{k}\), so \(G' = \langle \phi(g) \rangle\) is cyclic. Homomorphic images of cyclic groups are cyclic.The converse fails: the map from the Klein four-group \(\mathbf{Z}_{2} \times \mathbf{Z}_{2}\) onto \(\mathbf{Z}_{2}\) given by the first coordinate is surjective, G' is cyclic and G is not. The same example refutes (D) and the 'if and only if' in (C).

Q: Let G be a finite group and H, H' be two subgroups of G. Then :

The true statement is \(H = H' \Leftrightarrow aH = bH'\) for some \(a, b \in G\).If \(H = H'\), take \(a = b = e\). Conversely, if \(aH = bH'\), then \(H = a^{-1}bH'\). The left side contains \(e\), so the coset \(a^{-1}bH'\) contains \(e\), which means it is \(H'\) itself. Hence \(H = H'\).The other conditions are necessary but not sufficient. Different subgroups can have the same order and the same index, such as the three subgroups of order 2 in \(S_{3}\), so A and B fail. Normality says nothing about equality: in an abelian group every subgroup is normal, yet the subgroups can differ (D).

Q: Let G be a cyclic group of order 20. Then the number of elements of G of order 10 is :

There are 4 elements of order 10.A cyclic group of order \(n\) has exactly \(\varphi(d)\) elements of order \(d\) for each divisor \(d\) of \(n\). Here \(d = 10\) divides 20, and\[ \begin{aligned} \varphi(10) &= 10\left(1 - \tfrac{1}{2}\right)\left(1 - \tfrac{1}{5}\right) \\ &= 4 \end{aligned} \]Writing \(G = \langle g \rangle\), they are \(g^{2}, g^{6}, g^{14}, g^{18}\), the powers \(g^{k}\) with \(\gcd(k, 20) = 2\).10 is the order itself, not the number of elements having it. The elements of order 10 are exactly the generators of the unique subgroup of order 10, and a cyclic group of order 10 has \(\varphi(10) = 4\) generators, so 6 and 2 are wrong counts.

Q: The number of elements of order 3 in the direct product of groups \(\mathbf{Z}_{9} \times \mathbf{Z}_{3}\) is :

There are 8 elements of order 3.An element \((a, b)\) of \(\mathbf{Z}_{9} \times \mathbf{Z}_{3}\) satisfies \(3(a, b) = 0\) exactly when \(3a = 0\) in \(\mathbf{Z}_{9}\), i.e. \(a \in \{0, 3, 6\}\), and \(3b = 0\), which holds for all three \(b\). That gives \(3 \times 3 = 9\) elements of order dividing 3.Removing the identity leaves 8 elements of order exactly 3.9 forgets to remove the identity. 6 and 18 do not match the count of solutions of \(3(a, b) = 0\); 18 is in fact the number of elements of order 9.

Q: The subgroup lattice of a finite cyclic group is :

The subgroup lattice is a finite direct product of chains.A cyclic group of order \(n\) has exactly one subgroup of each order \(d\) dividing \(n\), and \(H_{d} \subseteq H_{e}\) exactly when \(d \mid e\). So its subgroup lattice is the lattice of divisors of \(n\). If \(n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}\), a divisor is chosen by picking an exponent \(0, \ldots, a_{i}\) for each prime independently, which makes the lattice the product of the chains \(\{0, 1, \ldots, a_{i}\}\).A product of chains is distributive, and every distributive lattice is modular, so 'modular but not distributive', 'not modular' and 'not distributive' are all false.

11.

Which of the following groups is not solvable ?

A.\(\mathbf{Z}_{150}\)
B.\(S_{4}\)
C.\(S_{5}\)
D.a non-abelian group of order 27

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The group that is not solvable is \(S_{5}\).

\(S_{5}\) contains the alternating group \(A_{5}\), which is simple and non-abelian. Its derived series is \(S_{5} \supset A_{5} \supset A_{5} \supset \cdots\), never reaching the identity, so \(S_{5}\) is not solvable. This is the group-theoretic reason the general quintic is not solvable by radicals.

The others are solvable. \(\mathbf{Z}_{150}\) is abelian. \(S_{4} \supset A_{4} \supset V_{4} \supset \{e\}\) has abelian factors. Every group of order 27 is a \(p\)-group, and every finite \(p\)-group is nilpotent, hence solvable, abelian or not.

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12.

Let H be a proper subgroup of the additive group (Q, +) of rational numbers. Then the quotient group Q/H must be :

A.finite of even order
B.finite of odd order
C.finite
D.infinite

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

Q/H must be infinite.

(Q, +) is divisible: for every \(q\) and every positive integer \(n\) there is \(r\) with \(nr = q\). A quotient of a divisible group is divisible, so Q/H is divisible. It is also non-trivial because H is proper.

A non-trivial finite group of order \(m\) cannot be divisible: if \(g\) is not the identity, there is no \(x\) with \(mx = g\), since \(mx = 0\) for every element. So Q/H cannot be finite, of either parity.

For example, with H = Z the quotient Q/Z is infinite: the classes of \(1/n\) for \(n = 1, 2, 3, \ldots\) are all different.

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13.

Let G be a group of order 61. Then the number of subgroups of G is :

A.61
B.2
C.7
D.1

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

G has 2 subgroups.

61 is prime, so by Lagrange's theorem a subgroup has order 1 or 61. The only subgroup of order 1 is {e}, and the only subgroup of order 61 is G itself. (A group of prime order is cyclic, generated by any non-identity element.)

61 counts the elements, not the subgroups; 1 forgets one of the trivial subgroups; and 7 has no connection with 61.

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14.

Let G be the group of invertible \(2 \times 2\) matrices over the field \(\mathbf{Z}_{2} = \{0, 1\}\). Then the number of elements in G is :

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The group GL(2, \(\mathbf{Z}_{2}\)) has 6 elements.

Choose the first column as any non-zero vector in \(\mathbf{Z}_{2}^{2}\): \(4 - 1 = 3\) ways. The second column must not be a multiple of the first, which rules out 0 and the first column itself: \(4 - 2 = 2\) ways. So

\[ \begin{aligned} |GL(2, \mathbf{Z}_{2})| &= (2^{2} - 1)(2^{2} - 2) \\ &= 3 \cdot 2 = 6 \end{aligned} \]

There are \(2^{4} = 16\) matrices in all, and 10 of them are singular, leaving 6. The options 2, 4 and 8 undercount; the group is in fact isomorphic to \(S_{3}\), acting on the three non-zero vectors.

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15.

The group \(\mathbf{Z}_{72}\) is the direct product of groups as :

A.\(\mathbf{Z}_{2} \times \mathbf{Z}_{2} \times \mathbf{Z}_{2}\) \(\times\ \mathbf{Z}_{3} \times \mathbf{Z}_{3}\)
B.\(\mathbf{Z}_{36} \times \mathbf{Z}_{36}\)
C.\(\mathbf{Z}_{9} \times \mathbf{Z}_{8}\)
D.\(\mathbf{Z}_{2} \times \mathbf{Z}_{36}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

\(\mathbf{Z}_{72} \cong\) \(\mathbf{Z}_{9} \times \mathbf{Z}_{8}\).

By the Chinese remainder theorem, \(\mathbf{Z}_{mn} \cong \mathbf{Z}_{m} \times \mathbf{Z}_{n}\) exactly when \(\gcd(m, n) = 1\). Here \(72 = 9 \times 8\) with \(\gcd(9, 8) = 1\).

The other products are not cyclic, because their factors share a common divisor: \(\mathbf{Z}_{2}^{3} \times \mathbf{Z}_{3}^{2}\) has exponent 6; \(\mathbf{Z}_{36} \times \mathbf{Z}_{36}\) has order 1296, not 72; and \(\mathbf{Z}_{2} \times \mathbf{Z}_{36}\) has order 72 but no element of order 72 (its exponent is 36).

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16.

Let \(\phi : G \to G'\) be a surjective group homomorphism. Then :

A.If G is cyclic, then G' is cyclic
B.If G' is cyclic, then G is cyclic
C.G is cyclic if and only if G' is cyclic
D.If G is not cyclic, then G' is not cyclic

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The correct statement is if G is cyclic, then G' is cyclic.

If \(G = \langle g \rangle\) and \(\phi\) is onto, every element of G' is \(\phi(g^{k}) = \phi(g)^{k}\), so \(G' = \langle \phi(g) \rangle\) is cyclic. Homomorphic images of cyclic groups are cyclic.

The converse fails: the map from the Klein four-group \(\mathbf{Z}_{2} \times \mathbf{Z}_{2}\) onto \(\mathbf{Z}_{2}\) given by the first coordinate is surjective, G' is cyclic and G is not. The same example refutes (D) and the 'if and only if' in (C).

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17.

Let G be a finite group and H, H' be two subgroups of G. Then :

A.\(H = H' \Leftrightarrow \#H = \#H'\)
B.\(H = H' \Leftrightarrow [G : H] = [G : H']\)
C.\(H = H' \Leftrightarrow aH = bH'\) for some \(a, b \in G\)
D.\(H = H' \Leftrightarrow\) both H and H' are normal in G

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The true statement is \(H = H' \Leftrightarrow aH = bH'\) for some \(a, b \in G\).

If \(H = H'\), take \(a = b = e\). Conversely, if \(aH = bH'\), then \(H = a^{-1}bH'\). The left side contains \(e\), so the coset \(a^{-1}bH'\) contains \(e\), which means it is \(H'\) itself. Hence \(H = H'\).

The other conditions are necessary but not sufficient. Different subgroups can have the same order and the same index, such as the three subgroups of order 2 in \(S_{3}\), so A and B fail. Normality says nothing about equality: in an abelian group every subgroup is normal, yet the subgroups can differ (D).

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18.

Let G be a cyclic group of order 20. Then the number of elements of G of order 10 is :

A.10
B.4
C.6
D.2

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

There are 4 elements of order 10.

A cyclic group of order \(n\) has exactly \(\varphi(d)\) elements of order \(d\) for each divisor \(d\) of \(n\). Here \(d = 10\) divides 20, and

\[ \begin{aligned} \varphi(10) &= 10\left(1 - \tfrac{1}{2}\right)\left(1 - \tfrac{1}{5}\right) \\ &= 4 \end{aligned} \]

Writing \(G = \langle g \rangle\), they are \(g^{2}, g^{6}, g^{14}, g^{18}\), the powers \(g^{k}\) with \(\gcd(k, 20) = 2\).

10 is the order itself, not the number of elements having it. The elements of order 10 are exactly the generators of the unique subgroup of order 10, and a cyclic group of order 10 has \(\varphi(10) = 4\) generators, so 6 and 2 are wrong counts.

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19.

The number of elements of order 3 in the direct product of groups \(\mathbf{Z}_{9} \times \mathbf{Z}_{3}\) is :

A.6
B.8
C.9
D.18

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

There are 8 elements of order 3.

An element \((a, b)\) of \(\mathbf{Z}_{9} \times \mathbf{Z}_{3}\) satisfies \(3(a, b) = 0\) exactly when \(3a = 0\) in \(\mathbf{Z}_{9}\), i.e. \(a \in \{0, 3, 6\}\), and \(3b = 0\), which holds for all three \(b\). That gives \(3 \times 3 = 9\) elements of order dividing 3.

Removing the identity leaves 8 elements of order exactly 3.

9 forgets to remove the identity. 6 and 18 do not match the count of solutions of \(3(a, b) = 0\); 18 is in fact the number of elements of order 9.

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20.

The subgroup lattice of a finite cyclic group is :

A.finite direct product of chains
B.modular but not distributive
C.not modular
D.not distributive

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The subgroup lattice is a finite direct product of chains.

A cyclic group of order \(n\) has exactly one subgroup of each order \(d\) dividing \(n\), and \(H_{d} \subseteq H_{e}\) exactly when \(d \mid e\). So its subgroup lattice is the lattice of divisors of \(n\). If \(n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}\), a divisor is chosen by picking an exponent \(0, \ldots, a_{i}\) for each prime independently, which makes the lattice the product of the chains \(\{0, 1, \ldots, a_{i}\}\).

A product of chains is distributive, and every distributive lattice is modular, so 'modular but not distributive', 'not modular' and 'not distributive' are all false.

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