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

Then

  1. $ {G/Z(G)} \cong C_6 $

  2. $ {G/Z(G)} \cong S_3 $

  3. $ {G/Z(G)} \cong C_2 $

  4. $ {G/Z(G)} \cong C_4 $

I am stuck on this problem. Can anyone help me please.... $ C_2,C_4,C_6.$ means what???......

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One certainly uses \cong to make the sign $\,\cong\,$, but we say the groups are isomorphic, not congruent. –  DonAntonio Jan 2 '13 at 15:08

3 Answers 3

up vote 2 down vote accepted

Since $M$ has ordedr 2 and $N$ has order 3, you can rule out $C_2,C_4$. And since $M,N$ don't commute, you can rule out $C_6$.

So as far as a technique for multiple choice, it's easy. Of course it's another matter to actually show the matrices generate a copy of $S_3$.

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@ coffeemath sir, $ C_2,C_4,C_6.$ means what???...... –  Prasanta Jan 2 '13 at 15:41
2  
Er...didn't you write these in your question?! Anyway, $\,C_n\,$ is a usual way to denote the cyclic group of order $\,n\,$ –  DonAntonio Jan 2 '13 at 15:53
    
@ DonAntonio sir,thanks......... –  Prasanta Jan 2 '13 at 17:49

As $MNM^{-1}=N^2$, the subgroup generated by $N$ is normal. This implies that $G$ is a semidirect product of $C_3$ by $C_4$. If $M^iN^j$ is central then, commuting with $M$, we see that $N^{2j}=N^j$. It follows that $j=0$. Now, commuting with $N$, it is easy to see that $i$ has to be even. This shows that $Z(G)=\{Id,M^2\}$, so that $G/Z(G)$ has order $6$. This quotient is not commutative and, since it has order $6$, it has to be isomorphic to $S_3$.

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thanks to all............ –  Prasanta Jan 2 '13 at 17:49

by taking example easily you can verify that the given group is not abelian. so in the option the only non-abelian group is $S_3$

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