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Thread: 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 $\displaystyle H$ is a normal subgroup of $\displaystyle GL(n,R)$, where

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

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

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

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

    so let $\displaystyle X \in AHA^{-1}$

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

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

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

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

    Does this effectively show that $\displaystyle H \triangleleft G$ (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 $\displaystyle \{ aH : a \in GL(n,R) \}$
    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 $\displaystyle G/H$, 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 $\displaystyle 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 $\displaystyle \mathbb{Z}_2$.
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