# Thread: Quotient Groups - Matrices

1. ## Quotient Groups - Matrices

Let $G=GL_2(Q,*)$ , H={ $\begin{pmatrix} 1&a\\ 0&1 \end{pmatrix} : a\in Q$ } a subgroup of G.

Find [G:H].

How do I do this?! I don't have any idea how to even look at it

Can you please explain to me what is this?

Thanks!

2. Two elements $A,B \in G$ are equivalent modulo $H$ iff
$AB^{-1} \in H$
<=>
there exists $C\in H$ , so that
$A=CB$
If you now write the matrices C and B explicit and calculate CB, you can see that in order for this equation to hold, B and A must have identical second rows.
So you can conclude that if A and B are equivalent, then they must have identical second rows at least or conversely: If the second rows of A and B are different, then A and B are not equivalent.
Since there are infinitely many matrices in G (i.e. invertible matrices with rational entries) which have pairwise different second rows, there are infinitely many elements that are all pairwise not equivalent and so the index [G:H] must be infinite.

3. Originally Posted by Iondor
Two elements $A,B \in G$ are equivalent modulo $H$ iff
$AB^{-1} \in H$ <-- Do you mean that A,B in G are in H iff ...?
<=>
there exists $C\in H$ , so that
$A=CB$
If you now write the matrices C and B explicit and calculate CB, you can see that in order for this equation to hold, B and A must have identical second rows. <-- if A,B are in H, then their second row is 0,1 anyway :O
So you can conclude that if A and B are equivalent, then they must have identical second rows at least or conversely: If the second rows of A and B are different, then A and B are not equivalent.
Since there are infinitely many matrices in G (i.e. invertible matrices with rational entries) which have pairwise different second rows, there are infinitely many elements that are all pairwise not equivalent and so the index [G:H] must be infinite.
I believe there's a misunderstanding here - GL2 is the general linear group of all invertable matrices of size 2x2. What do you mean by "A,B in G are equivalent modulo H"? How can 2 matrices be equivalent?

Thanks

4. Originally Posted by adam63
I believe there's a misunderstanding here - GL2 is the general linear group of all invertable matrices of size 2x2. What do you mean by "A,B in G are equivalent modulo H"? How can 2 matrices be equivalent?

Thanks
$A$, $B$ are equivalent modulo $H$ means that $AH = BH$ in $G/H$, i.e. they are in the same coset. It can be shown that two elements $A, B$ are in the same coset if and only if $AB^{-1}\in H$.

5. Aha!

Now I understand it, thank you very much !