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Math Help - Normal Subgroup of the General Linear Group?

  1. #1
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    Normal Subgroup of the General Linear Group?

    Hello,

    I am trying to show that the group H is a normal subgroup of GL(n,R), where

    H = \{A \in GL(n,R) : det(A) > 0\}

    I am just wondering if this proof is in the correct style (Normal Subgroup Test).

    So, let A be an arbitrary matrix in GL(n,R)
    then, A^{-1} is also in GL(n,R)

    now, we want to show AHA^{-1} \subseteq H,\  \forall A \in GL(n,R) to show that H \triangleleft G (H is normal in G). This is the normal subgroup test.

    so let X \in AHA^{-1}

    then X = AH_oA^{-1} for some H_o \in H

    then det(X) = det(AH_oA^{-1}) = det(A)det(H_o)det(A^{-1}) = det(H_o) > 0 since H_o \in H where H = \{A \in GL(n,R) : det(A) > 0\}

    Since det(X) > 0, then X \in H

    So X \in AHA^{-1} \rightarrow X \in H
    so AHA^{-1} \subseteq H\  \forall A \in GL(n,R) since A was arbitrary!

    Does this effectively show that H \triangleleft G (H is normal in G)?

    Thank you!!

    ----
    Also, if the above is correct,
    I am having trouble with the second part of the question, which asks:

    Which common, known group is isomorphic to the factor group G/H for the groups G,H in the above question?

    (the factor group G/H is defined as the group of left-cosets \{ aH : a \in GL(n,R) \}
    I can't see right away that the factor group G/H is isomorphic to anything that I could think of!

    Thanks!!
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  2. #2
    Senior Member roninpro's Avatar
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    Yes, your argument is correct.

    To identify G/H, look carefully at your cosets. I think that there are only two.
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  3. #3
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    Are the two cosets,
    matrices with positive nonzero determinant
    and matrices with negative nonzero determinant?

    Would that possibly be isomorphic to  Z_2 under addition?

    I feel like this can't be the case...

    Thank you for confirming the normal subgroup test. I thought I understood it in class until I tried applying it, then I was unsure! Thanks.
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  4. #4
    Senior Member roninpro's Avatar
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    Yes, those are the two cosets.

    You can prove via multiplication tables that any group of order two is really \mathbb{Z}_2.
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