Discussion
Let \(GL_{2}(\mathbb{C})\) denote the group of \(2 \times 2\) invertible complex matrices with usual matrix multiplication. For \(S, T \in GL_{2}(\mathbb{C})\), \(\langle S, T\rangle\) denotes the subgroup generated by S and T. Let \(S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \in GL_{2}(\mathbb{C})\) and \(G_{1}, G_{2}, G_{3}\) be three subgroups of \(GL_{2}(\mathbb{C})\) given by
\(G_{1} = \langle S, T_{1}\rangle\), where \(T_{1} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}\),
\(G_{2} = \langle S, T_{2}\rangle\), where \(T_{2} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}\),
\(G_{3} = \langle S, T_{3}\rangle\), where \(T_{3} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\).
Let \(Z(G_{i})\) denote the center of \(G_{i}\) for i = 1, 2, 3.
Which of the following statements is correct?
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.
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