# Normal Subgroup of the General Linear Group?

• November 19th 2010, 08:55 AM
matt.qmar
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!!
• November 19th 2010, 09:31 AM
roninpro

To identify $G/H$, look carefully at your cosets. I think that there are only two.
• November 19th 2010, 11:22 AM
matt.qmar
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.
• November 19th 2010, 07:39 PM
roninpro
Yes, those are the two cosets.

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