Let $p$ is odd prime. For each $1\leqslant d\leqslant p-1.$ Let $SL^d_n(\mathbb{F}_p) = \{X\in GL_n(\mathbb{F}_p)\,\,:\,\,(det X)^d = 1\}$ be the special linear group of degree $d$ over the finite field $\mathbb{F}_p.$
Show that $SL^{p-1}_n(\mathbb{F}_p) = GL_n(\mathbb{F}_p)?$ Here's what I have:
Proof. We have homomorphism $\det: GL_n(\mathbb{F}_p)\to \mathbb{F}_p^* = \mathbb{F}_p\setminus\{0\}$ is an epimorphism.
For all $X\in GL_n(\mathbb{F}_p),$ then $\det X\in \mathbb{F}_p^*,$ so $(\det X, p) = 1.$
Hence, from Fermat's little theorem, one gets $(det X)^{p-1}\equiv 1 \pmod p.$
Therefore, $ SL^{p-1}_n(\mathbb{F}_p) = GL_n(\mathbb{F}_p).$
Is it right or wrong?