Show that $GL_n(\mathbb{F}_{23})$ has subgroup of index $2$

Question

Show that $GL_n(\mathbb{F}_{23})$ has subgroup of index $2$

We know that $|GL_n(\mathbb{F}_{23})|=(23^n-1)(23^n-23) \dots (23^n-23^{n-1})$

and therefore is even.

I thought about Cauchy theorem but I don't know that $GL_n(\mathbb{F}_{23})$ is abelian… Any ideas?

EDIT

Ok so I get it that I should is the first isomorphism theorem, and get that the kernel is of index $2$, but I don't understand how to build that isomorphism…

Answer 1

$\det: GL_n(\Bbb F_{23})\to\Bbb F_{23}^*$ is a surjective homomorphism. Now, $\Bbb F_{23}^*\cong \Bbb Z/22\Bbb Z$, thus it surjects onto $\Bbb Z/2\Bbb Z$ via the quotient map $\pi:\Bbb Z/22\Bbb Z\to\frac{\Bbb Z/22\Bbb Z}{2\Bbb Z/22\Bbb Z}$. So, $\ker(\pi\circ\det)$ has index $2$.