## The special linear group is a subgroup of the general linear group

Let $F$ be a field and define $SL_n(F) \subseteq GL_n(F)$ by $SL_n(F) = \{ A \in GL_n(F) \ |\ \mathsf{det}(A) = 1 \}$. Prove that $SL_n(F) \leq GL_n(F)$.

We assume that for all $A, B \in GL_n(F)$, $\mathsf{det}(AB) = \mathsf{det}(A) \mathsf{det}(B)$. From this we deduce that if $A \in GL_n(F)$, then $1 = \mathsf{det}(AA^{-1}) = \mathsf{det}(A) \mathsf{det}(A^{-1})$, so that $\mathsf{det}(A^{-1}) = \mathsf{det}(A)^{-1}$.

Note that $SL_n(F)$ is not empty since $I_n \in SL_n(F)$. Now let $A, B \in SL_n(F)$. Then we have $\mathsf{det}(AB^{-1}) = \mathsf{det}(A) \mathsf{det}(B)^{-1} = 1$, so that $AB^{-1} \in SL_n(F)$. By the subgroup criterion, then, $SL_n(F) \leq GL_n(F)$.