**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.1 Exercise 2.1.9**

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 $SL_n(F) \leq GL_n(F)$.

Solution: For all $A, B \in GL_n(F)$, we have $\mathsf{det}(AB) = \mathsf{det}(A) \mathsf{det}(B)$.

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, $SL_n(F) \leq GL_n(F)$.