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Consider the subgroup $G$ of $GL_{2}(\mathbb{C})$ generated by $A=\begin{pmatrix} \omega & 0 \\ 0 & \omega^{2} \end{pmatrix}$ and $B=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$ where $\omega=e^{\frac{2\pi i}{3}}$. Is there an isomorphism between $G$ and $H:=\langle a\in A,b\in B|a^{6}=I,b^{2}=a^{3}=(ab)^{2}\rangle$?

I computed that $A^3=B^4=I$ so is this enough so prove that $G$ is of order $12$? And I have very little intuition how to show the isomorphism. Probably by computing its 2-Sylow subgroup and whether it is a normal subgroup?

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    Looks like the problem is from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=500531&p2816667#p28166672012-10-06

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The elements of the group are $I, A, A^2, B, B^2,B^3, AB,AB^2, AB^3, A^2B, A^2 B^2, A^2B^3$. Use the relation $BA=A^2 B$ to show that all other products reduce to these 12.

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    o.k. So we agree that $G$ is the "strange" group of order 12!2012-10-07