Prove that SL_n(F) is a subgroup of GL_n(F)

Let FF be any field. Prove that SLn(F)SL_n(F) is a subgroup of GLn(F)GL_n(F).

Proof. Let FF be any field. Recall that the general linear group is defined as:
GLn(F)={A  A is an n×n invertible matrix with entries from F}\begin{align*} GL_n(F) = \{A \ | \ A \text{ is an } n \times n \text{ invertible matrix with entries from } F\} \end{align*}
and the special linear group is defined as:
SLn(F)={AGLn(F)  det(A)=1}.\begin{align*} SL_n(F) = \{A \in GL_n(F) \ | \ det(A) = 1\}. \end{align*}
We want to show that if A,BSLn(F)A,B \in SL_n(F), then A1SLn(F)A^{-1} \in SL_n(F) and ABSLn(F)AB \in SL_n(F). Assume for the first case that ASLn(F)A \in SL_n(F). Since det(A)=1det(A) = 1, and therefore inverse, we have that:
det(A1)=1det(A)=11=1.\begin{align*} det(A^{-1}) = \frac{1}{det(A)} = \frac{1}{1} = 1. \end{align*}
So A1SLn(F)A^{-1} \in SL_n(F). For the second case, assume that A,BSLn(F)A,B \in SL_n(F). Then det(A)=det(B)=1det(A) = det(B) = 1. Now we have that:
det(AB)=det(A)det(B)=1.\begin{align*} det(AB) = det(A)det(B) = 1. \end{align*}
Therefore, ABSLn(F)AB \in SL_n(F).

So SLn(F)SL_n(F) is a subgroup of GLn(F)GL_n(F).

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