Groups and Subgroups Questions and Answers

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.

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.

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.

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.

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).

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).

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).

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.

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.

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.