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).

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