Discussion
Let \(\omega = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\), \(M = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\), \(N = \begin{pmatrix} \omega & 0 \\ 0 & \omega^{2} \end{pmatrix}\) and \(G = \langle M, N \rangle\) be the group generated by the matrices \(M\) and \(N\) under matrix multiplication. Then
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.
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