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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 $$? Is it intended to mean that $a$ should be the correspondent of $A$ and $b$ of $B$?2012-10-05
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    Yes. This is what I meant.2012-10-05
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    You also have $BA=A^{-1} B=A^2 B$. This should help you determine all the elements of $G$.2012-10-05
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    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." http://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).2012-10-05
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    Sorry if my language is really informal or even incorrect; I'm not too comfortable with group theory.2012-10-05
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    Your second statement in the definition of $H$ seems false. If $a=\omega$ and $b=i$, $b^2\ne a^3$. Likewise, if $a=i$ and $b=\omega$, $b^2\ne a^3$. I can see the statement only being true if $a=b=0$. Can you explain this?2012-10-05
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    But $a\ne\omega$ as $\omega$ is a complex number, $a$ is a matrix. I just took those matrices that satisfy the relations in given in $H$'s definition.2012-10-05
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    Sorry if this is a stupid question, but: If $a$ is a matrix, then how is it true that $a^6=1$? Do you mean that $a^6=I$, where $I$ is the identity matrix? I am quite confused by your notation, especially $a \in A$ and $b \in B$. So, what do $a \in A$ and $b \in B$ mean precisely?2012-10-05
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    Indeed, a bad notation. I'm trying to understand the notation behind the problem even myself.2012-10-05
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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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    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?2012-10-05
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    I take my previous comment back. Hm.2012-10-05
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    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 $b^4=1$?2012-10-05
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    @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.2012-10-05
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    Yes. But there is an extra relation $BA=A^2B$. It allows me to switch $B$ with $A$. I think this group has no element of order 6. It is probably $A_4$ the alternating group of order 12.2012-10-05
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    So I have to check that for any element $x$, either $x^6\ne I$ or ($x^6=I$ and ($x^3=I$ or $x^2=I$)). Is it easier to show that the 3-Sylow subgroup is not normal.2012-10-05
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    G and H have order 12. Find an element of order 6 in G if you can.2012-10-05
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    It definitely is not $A_4$, as the subgroup generated by $A$ is normal. $AB^2$ is an element of order $6$.2012-10-05
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    There are three non-abelian groups of order 12. If it is not $A_4$ then it is either the dihedral group $D_6$ or a subgroup of $S_6 \times Z_4$ of order 12 (generated by elements $a$, $b$ such that $|a|=6$, $b^2=a^3$ and $ba=a^{-1} b$. $T$ is called the binary dihedral group. It should have a unique element of order 2.2012-10-06
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    Hint: it is the dihedral group. Think about how both $A$ and $B$ act on this element $AB^2$ of order 6.2012-10-07
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    @Steve D But if it was the dihedral group then it should have 7 elements of order 2, i.e. the six reflections and the rotation to 3rd power. In this group, I can only see $B^2$ having order 2. Am I right?2012-10-07
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    @PantelisDamianou: whoops! I meant *not* the dihedral group. You can see this as $B$ is an element of order $4$ sitting outside what would be the usual "cyclic subgroup".2012-10-07
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    o.k. So we agree that $G$ is the "strange" group of order 12!2012-10-07