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**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.