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!!

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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!!

Yes, your argument is correct.

To identify $\displaystyle G/H$, look carefully at your cosets. I think that there are only two.

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.

Yes, those are the two cosets.

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