Show that $\operatorname{GL}_n(\mathbb{R})/\operatorname{GL}_n^+(\mathbb{R}) \cong \mathbb{Z}_2$

Question

Here, $\operatorname{GL}_n(\mathbb{R})$ denotes the standard general linear group and $\operatorname{GL}_n^+(\mathbb{R})$ is the set of all $n \times n$ matrices with real entries and positive determinant.

An earlier exercise requires that we show that $\operatorname{GL}_n^+(\mathbb{R})$ is a normal subgroup of $\operatorname{GL}_n(\mathbb{R})$, so I assume this problem requires the use of the First Isomorphism Theorem. The thing is, though, I just don't understand how to apply it.

My guess is that we can show that $\operatorname{GL}_n(\mathbb{R})/\operatorname{GL}_n^+(\mathbb{R}) \cong \mathbb{Z}_2$ by constructing a homomorphism from $\operatorname{GL}_n(\mathbb{R})$ to $\mathbb{Z}_2$ whose kernel is $\operatorname{GL}_n^+(\mathbb{R})$, but I'm missing anything with the image.

How would you go about using the FIT to prove this? In addition, and more generally, how does one use the FIT to prove that $G/H \cong J$, where $G, H, J$ are groups.

Answer 1

Hint: show that $\phi:GL_n(\Bbb R) \to \{-1,1\}$, given by:

$\phi(A) = \dfrac{\det(A)}{|\det(A)|}$

is a surjective group homomorphism, and observe the image is isomorphic to $\Bbb Z_2$.

Answer 2

Of course we can construct a mapping directly to $\mathbb{Z_{2}}$.

Let $\varphi: GL_{n}(\mathbb{R})\to \mathbb{Z_{2}}$; $\varphi(A)=[0]$ if $A\in GL_{n}^{+}(\mathbb{R})$ and $\varphi(A)=[1]$ in other cases. We know that $\det(AB)=\det A\det B$, so it's easy to check that $\varphi(AB)=\varphi(A)+\varphi(B)$, and therfore $\varphi$ is a surjective homomorphism from $GL_{n}(\mathbb{R})$ to $\mathbb{Z_{2}}$. The kernel of this homomorphism, of course is $GL_{n}^{+}(\mathbb{R})$.