[proofplan] We first fix notation by identifying the determinant map in the statement with a named homomorphism $\delta:GL_n(k)\to k^\times$. Its kernel is then computed directly from the definition of a kernel and the definition of $SL_n(k)$ as the determinant-one subgroup of $GL_n(k)$. Finally, instead of merely citing the normal-kernel theorem, we prove normality directly by conjugating an arbitrary determinant-one matrix inside $GL_n(k)$ and using the homomorphism property of $\delta$. [/proofplan] [step:Name the determinant homomorphism] Let \begin{align*} \delta:GL_n(k)\to k^\times \end{align*} denote the determinant homomorphism from the statement, so $\delta(A)=\det A$ for every $A\in GL_n(k)$. Thus $\delta$ is a [group homomorphism](/page/Group%20Homomorphism) and its kernel is defined. [guided] The statement declares the determinant as a homomorphism \begin{align*} \det:GL_n(k)\to k^\times. \end{align*} We introduce the shorter name $\delta$ for this same map: \begin{align*} \delta:GL_n(k)\to k^\times. \end{align*} For every matrix $A\in GL_n(k)$, this means $\delta(A)=\det A$. Since $\delta$ is the determinant homomorphism named in the statement, it is a group homomorphism from $GL_n(k)$ to $k^\times$, and therefore the subgroup $\ker\delta$ is meaningful. [/guided] [/step] [step:Identify the kernel with the determinant-one subgroup] By the definition of the [kernel](/page/Kernel) of the homomorphism $\delta$, \begin{align*} \ker\delta=\{A\in GL_n(k):\delta(A)=1\}. \end{align*} Since $\delta(A)=\det A$, this becomes \begin{align*} \ker\delta=\{A\in GL_n(k):\det A=1\}. \end{align*} By the definition of the special linear group, \begin{align*} SL_n(k)=\{A\in GL_n(k):\det A=1\}. \end{align*} Therefore \begin{align*} SL_n(k)=\ker\delta=\ker(\det:GL_n(k)\to k^\times). \end{align*} [/step] [step:Prove normality by conjugating inside $GL_n(k)$] Let $S:=SL_n(k)$. To prove $S\trianglelefteq GL_n(k)$, let $B\in GL_n(k)$ and let $A\in S$. Since $A\in S$, we have $\delta(A)=\det A=1$. Since $B\in GL_n(k)$, the inverse $B^{-1}$ exists in $GL_n(k)$. Because $\delta$ is a group homomorphism, \begin{align*} \delta(BAB^{-1})=\delta(B)\delta(A)\delta(B^{-1}). \end{align*} Using $\delta(A)=1$ and $\delta(B^{-1})=\delta(B)^{-1}$ in the group $k^\times$, we obtain \begin{align*} \delta(BAB^{-1})=1. \end{align*} Thus $BAB^{-1}\in SL_n(k)$ for every $A\in SL_n(k)$ and every $B\in GL_n(k)$. Hence \begin{align*} SL_n(k)\trianglelefteq GL_n(k). \end{align*} [/step]