Exercise 2.1.9 (Special linear group)

Proof.

• ${\mathrm{SL}}_n(F)\subseteq {\mathrm{GL}}_n(K)$ and $\det <!--\; FUNCTION\; APPLICATION\; -->(I_n)=1$, thus $I_n\in {\mathrm{SL}}_n(F)$, so ${\mathrm{SL}}_n(F)\ne \varnothing$.
• If $A,B\in {\mathrm{SL}}_n(F)$, then $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\det <!--\; FUNCTION\; APPLICATION\; -->(B)=1$, thus $\det <!--\; FUNCTION\; APPLICATION\; -->(A{B}^{-1})=\det <!--\; FUNCTION\; APPLICATION\; -->(A)∕\det <!--\; FUNCTION\; APPLICATION\; -->(B)=1$, thus $A{B}^{-1}\in {\mathrm{SL}}_n(F)$.

${\mathrm{SL}}_n(F)$ is a subgroup of ${\mathrm{GL}}_n(F)$. □