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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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How did you get the presented group $<a,b|...>$? Is it intended to mean that $a$ should be the correspondent of $A$ and $b$ of $B$? –  Berci Oct 5 '12 at 14:00
    
Yes. This is what I meant. –  Student Oct 5 '12 at 14:01
    
You also have $BA=A^{-1} B=A^2 B$. This should help you determine all the elements of $G$. –  PAD Oct 5 '12 at 14:01
    
I would enumerate all of the group elements and then employ this theorem: "[. . .]if a finite group is generated by a subset S, then each group element may be expressed as a word from the alphabet S of length less than or equal to the order of the group." en.wikipedia.org/wiki/Generating_set_of_a_group If I'm understanding correctly, that means that elements like $ABA$ reduce down. So, you can say that you've exhaustively listed all of $G$ once you list all the elements that are composed of $A$ and $B$ and the powers of $A$ and $B$ themselves (along with the identity). –  000 Oct 5 '12 at 14:04
    
Sorry if my language is really informal or even incorrect; I'm not too comfortable with group theory. –  000 Oct 5 '12 at 14:05
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up vote 2 down vote accepted

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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Are you working toward addressing the isomorphism? Maybe I'm wrong, but it would seem the order of the second group is not the same as the order of the first group. Hence, there is not an isomorphism. What says you? –  000 Oct 5 '12 at 14:12
    
I take my previous comment back. Hm. –  000 Oct 5 '12 at 14:36
    
Why would knowing that a and b have order 3 and 4 imply that they generate a group of order 12? How would you simplify ababababababababa just knowing $a^3 = 1$ and $b^4=1$? –  Noah Snyder Oct 5 '12 at 14:46
    
@Student A group generated by two torsion elements need not itself have finite order. There is no guarantees that in products of powers of the two elements, one power may "move through" the other and change into something else: it is this fact that allows us to limit the number of products of the two elements that are possible by their respective orders. –  anon Oct 5 '12 at 14:47
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o.k. So we agree that $G$ is the "strange" group of order 12! –  PAD Oct 7 '12 at 22:45
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