Let $k$ be a field and let $n\in\mathbb{N}$. Let $M_n(k)$ denote the ring of $n\times n$ matrices over $k$, let $I_n\in M_n(k)$ denote the identity matrix, let $GL_n(k)$ denote the group of invertible elements of $M_n(k)$ under matrix multiplication, and let $k^\times$ denote the multiplicative group of nonzero elements of $k$. Let

\begin{align*} \det:GL_n(k)\to k^\times \end{align*}

denote the determinant homomorphism, and define the special linear group by

\begin{align*} SL_n(k)=\{A\in GL_n(k):\det A=1\}. \end{align*}

Then

\begin{align*} SL_n(k)=\ker(\det:GL_n(k)\to k^\times). \end{align*}

Consequently,

\begin{align*} SL_n(k)\trianglelefteq GL_n(k). \end{align*}