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Math Help - Quotient Groups - Matrices

  1. #1
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    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!
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Iondor View Post
    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
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam63 View Post
    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.
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  5. #5
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    Aha!

    Now I understand it, thank you very much !
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