Is a subgroup of GL_2(C) a group of order 12?

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

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.
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
o.k. So we agree that $G$ is the "strange" group of order 12! –  PAD Oct 7 '12 at 22:45