1. Quotient Groups - Matrices

Let $\displaystyle G=GL_2(Q,*)$ , H={ $\displaystyle \begin{pmatrix} 1&a\\ 0&1 \end{pmatrix} : a\in Q$ } 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 $\displaystyle A,B \in G$ are equivalent modulo $\displaystyle H$ iff
$\displaystyle AB^{-1} \in H$
<=>
there exists $\displaystyle C\in H$ , so that
$\displaystyle A=CB$
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 $\displaystyle A,B \in G$ are equivalent modulo $\displaystyle H$ iff
$\displaystyle AB^{-1} \in H$ <-- Do you mean that A,B in G are in H iff ...?
<=>
there exists $\displaystyle C\in H$ , so that
$\displaystyle A=CB$
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

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