Groups and Subgroups Questions and Answers

Answer: B

The quotient is \(G/Z(G) \cong S_{3}\).

Compute the relations. \(M^{2} = -I\), so \(M\) has order 4, and \(N^{3} = I\) since \(\omega^{3} = 1\). Using \(M^{-1} = -M\):

\[ MNM^{-1} = \begin{pmatrix} \omega^{2} & 0 \\ 0 & \omega \end{pmatrix} = N^{-1}. \]

So \(G\) is the dicyclic group of order 12: the elements \(M^{a}N^{b}\) with \(a = 0, 1, 2, 3\) and \(b = 0, 1, 2\). Its centre is \(\{I, -I\} = \langle M^{2} \rangle\), because \(-I\) commutes with everything while \(M\) and \(N\) do not commute.

Hence \(G/Z(G)\) has order 6, generated by the images of \(N\) (order 3) and \(M\) (order 2), with \(MNM^{-1} = N^{-1}\). That is the dihedral group of order 6, which is \(S_{3}\). It is non-abelian, so it is not \(C_{6}\); and orders 2 and 4 are wrong for a quotient of order 6.

Answer: B

The smallest non-trivial such group has order 2.

The relation \(yx = x^{4}y\) says \(yxy^{-1} = x^{4}\). Conjugating twice and using \(y^{2} = e\):

\[ \begin{aligned} x &= y^{2}xy^{-2} = (x^{4})^{4} = x^{16} \\ x^{15} &= e \end{aligned} \]

Together with \(x^{7} = e\), and \(\gcd(15, 7) = 1\), this forces \(x = e\). So the group is generated by \(y\) alone with \(y^{2} = e\), and the smallest non-trivial example is the cyclic group of order 2 (\(x = e\), \(y\) of order 2), which satisfies every relation.

Order 1 is the trivial group, which the question excludes. Orders 7 and 14 would need \(x\) of order 7, but \(x \mapsto x^{4}\) has order 3 as an automorphism of \(C_{7}\), so it cannot be conjugation by an element of order 2.

Answer: B

The centre has 2 elements.

The relations \(o(x) = 4\), \(o(y) = 2\) and \(xy = yx^{3} = yx^{-1}\) describe the dihedral group of order 8, the symmetries of a square, with \(x\) a rotation and \(y\) a reflection.

An element \(x^{k}\) commutes with \(y\) only if \(x^{k} = x^{-k}\), that is \(k = 0\) or 2. No element \(x^{k}y\) commutes with \(x\), since \(x \cdot x^{k}y = x^{k+1}y\) while \(x^{k}y \cdot x = x^{k-1}y\). So \(Z(G) = \{e, x^{2}\}\).

8 would need G to be abelian, which it is not; a centre of order 4 would make \(G/Z(G)\) cyclic of order 2, forcing G to be abelian; and a non-trivial 2-group cannot have centre of order 1.

Answer: D

The correct statement is \(Z(G_{2})\) is isomorphic to \(Z(G_{3})\).

S has order 4 with \(S^{2} = -I\).

• \(G_{1}\): \(T_{1} = iI\) is scalar and commutes with S, so \(G_{1}\) is abelian of order 8 (\(\cong \mathbb{Z}_{4} \times \mathbb{Z}_{2}\)), and \(Z(G_{1}) = G_{1}\).
• \(G_{2}\): \(T_{2}^{2} = -I = S^{2}\) and \(ST_{2} = -T_{2}S\), the relations of the quaternion group \(Q_{8}\), whose centre is \(\{\pm I\}\).
• \(G_{3}\): \(T_{3}\) is a reflection with \(T_{3}ST_{3} = S^{-1}\), so \(G_{3}\) is the dihedral group of order 8, whose centre is also \(\{\pm I\}\).

Both centres are cyclic of order 2, hence isomorphic. \(G_{1}\) is abelian and \(G_{3}\) is not, so they are not isomorphic; \(Z(G_{1})\) has order 8, not 2; and \(Z(G_{3})\) contains -I, so it is not trivial.

Answer: D

Both I and II are FALSE.

\((\mathbb{Q}, +)\) is divisible: every element can be divided by every positive integer. Any quotient of a divisible group is divisible.

I: a finite divisible group is trivial, since if \(|Q/G| = n\) then every element equals \(n\) times some element, which is 0. So \(\mathbb{Q}/G\) finite forces \(G = \mathbb{Q}\), not a proper subgroup.

II: \((\mathbb{Z}, +)\) is not divisible (1 is not twice an integer), so it cannot be a quotient of \(\mathbb{Q}\). Both statements fail.

Answer: D

The FALSE statement is \(\mathbb{Z}_{6}\) is isomorphic to a quotient group of \(S_{4}\).

The normal subgroups of \(S_{4}\) are \(\{e\}\), the Klein four-group \(V_4\), \(A_4\) and \(S_4\). Their quotients have orders 24, 6, 2 and 1, and are \(S_4\), \(S_4/V_4 \cong S_3\), \(\mathbb{Z}_2\) and the trivial group. The only quotient of order 6 is \(S_3\), which is non-abelian, so the cyclic group \(\mathbb{Z}_6\) is not a quotient.

The other statements are true. The permutations fixing the symbol 4 form a subgroup isomorphic to \(S_3\). The 3-cycle \((1\,2\,3)\) generates a subgroup isomorphic to \(\mathbb{Z}_3\). And \(S_4/V_4 \cong S_3\), as above, so \(S_3\) is also a quotient.

Answer: C

The FALSE statement is: for all non-zero \(\alpha: \mathbb{Z} \to G\) and \(\beta: G \to K\), \(\beta \circ \alpha\) is non-zero.

Pick \(g \in G\) with \(f(g) = 1\) and define \(\phi(n) = ng\). Then \(f \circ \phi\) is the identity, so the first statement is true. Since \(G\) is abelian, every \(x\) splits as \((x - \phi(f(x))) + \phi(f(x))\), giving \(G = K \oplus \phi(\mathbb{Z})\). Then \(\phi\) is injective, so \(\phi(\mathbb{Z}) \cong \mathbb{Z}\) is a subgroup of \(G\), and \(G/\phi(\mathbb{Z}) \cong K\), so \(K\) is a quotient of \(G\). Both of those statements are true.

The composition statement fails in general. Take \(G = \mathbb{Z} \oplus \mathbb{Z}\) with \(f(a,b) = b\), so \(K = \mathbb{Z} \oplus 0\). Let \(\alpha(n) = (0,n)\) and \(\beta(a,b) = (a,0)\). Both are non-zero, yet \(\beta \circ \alpha = 0\). The statement can hold only in the degenerate case \(K = 0\), where no non-zero \(\beta\) exists, so it is not true for such groups in general.

Answer: B

The TRUE statement is: \((G, \Delta)\) is an abelian group but not cyclic.

Symmetric difference is associative and commutative. The empty set is the identity, since \(A \,\Delta\, \emptyset = A\), and every set is its own inverse, since \(A \,\Delta\, A = \emptyset\). So \((G, \Delta)\) is an abelian group of order \(2^{4} = 16\) in which every non-identity element has order 2; it is isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2\).

A cyclic group of order 16 would need an element of order 16, so the group is not cyclic. For the same reason there is no element of order 4 or 8. \(\{1\}\) is not the identity: \(\{1\} \,\Delta\, \{1\} = \emptyset \neq \{1\}\), and \(\{1\} \,\Delta\, \{2\} = \{1,2\}\), not \(\{2\}\).

Answer: D

The true statement is the order of a is finite and divides n.

Since \(a^{n} = e\), some positive power of a is the identity, so its order k (the least such power) is finite. Divide: \(n = qk + r\) with \(0 \le r \lt k\). Then \(a^{r} = a^{n}(a^{k})^{-q} = e\), and minimality of k forces \(r = 0\), so k divides n.

The others need not hold. G can be infinite: in the circle group, \(a = -1\) satisfies \(a^{2} = 1\). The order of a can be a proper divisor of n: the identity itself satisfies \(e^{n} = e\) for every n. And a subgroup of order n need not contain a: in \(\mathbf{Z}_{2} \times \mathbf{Z}_{2}\), the element (1, 0) has square equal to the identity but is not in the subgroup {(0, 0), (0, 1)} of order 2.

Answer: A

The true statement is if the cardinality of H is odd then n is odd.

H consists of the identity together with the elements of order 2. The other elements of G pair off with their inverses, \(a \ne a^{-1}\), so

\[ n = |H| + 2 \times (\text{number of pairs}) \]

Hence \(|H|\) and n have the same parity. If \(|H|\) is odd, n is odd; and if n is even, \(|H|\) is even, which also shows the statement 'if \(|H|\) is even then n is odd' is false.

The other two fail on small examples. In \(S_{3}\) (order 6), H is the identity and the three transpositions, so \(|H| = 4\), which does not divide 6. In the dihedral group of order 8, H has the identity and five elements of order 2, so \(|H| = 6\), not a power of two.