# Matrix groups, GL_2(C)

1. Sep 24, 2011

### burak100

1. The problem statement, all variables and given/known data

$A= \left( \begin{matrix} i & 0 \\ 0 &-i \end{matrix} \right)$
, $B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$
\\
Show that $\langle A, B \rangle$ is subgroup of $GL_2(\mathbb{C})$. And Show that $\langle A, B \rangle$ generated by $A$ and $B$, and order of $\langle A, B \rangle$ is 8 ?

2. Relevant equations

$GL_2(\mathbb{C}) = \big\lbrace X \in M_2(\mathbb{C}) ~~\vert ~~ \exists Y\in M_2(\mathbb{C}) ~ with~ XY=YX=I \big\rbrace$ \\
which $Y$ is inverse of $X$

3. The attempt at a solution

2. Sep 24, 2011

### Robert1986

And what have you done?? What do you have to do to show something is a subgroup?

3. Sep 24, 2011

### burak100

$\langle A, B \rangle = \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix} \right)$ and det(<A,B>)=-1, hence det(<A,B>) in $GL_2(\mathbb{C})$. right?

on the other hand, if we want to show $\langle A, B \rangle$ generated by A and B,
we need to show that A and B are linear independent ???

Last edited: Sep 24, 2011
4. Sep 24, 2011

### burak100

I am confused, because
$\langle A, B \rangle$$= \Big\langle \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\Big\rangle$$= \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix} \right)$ right?

and this is just an element of $GL_2{\mathbb{C}}$, not a group of $GL_2{\mathbb{C}}$, right?????

Last edited: Sep 24, 2011
5. Sep 24, 2011

### Dick

Right. I'm not sure what <A,B> is supposed to mean, but I think you just supposed to check that the group generated by all possible products of A and B is a subgroup of order 8.

6. Sep 24, 2011

### burak100

So , should I try to find all possible products of A and B , or is there some trick to find it?

7. Sep 24, 2011

### Dick

Well, A^4=I and B^2=I, right? Showing AB=(-BA) would also help a lot.

8. Sep 25, 2011

### burak100

I try to calculate possibilities,

I, A, B, AB, BA, AAB, AAA, BAA

and there are 8 elements , is it the answer?

9. Sep 25, 2011

### Dick

The answer would be a PROOF that those 8 elements form a group. Just listing them isn't enough. Besides, I don't think all of those are different. Isn't AAB=BAA?