Special linear group is fully characteristic in general linear group

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Special linear group (?)) satisfying a particular subgroup property (namely, Fully characteristic subgroup (?)) in a particular group or type of group (namely, General linear group (?)).

Statement

Let $k$ be a field and $n$ be a natural number. Let $GL_n(k)$ denote the group of invertible $n \times n$ matrices over $k$ under multiplication and $SL_n(k)$ denote the subgroup comprising matrices of determinant one. $SL_n(k)$ is a fully characteristic subgroup of $GL_n(k)$: every endomorphism of $GL_n(k)$ restricts to an endomorphism of $SL_n(k)$.

Facts used

1. Commutator subgroup of general linear group is special linear group: This is true except in the case where $n = 2$ and the field $k$ has exactly two elements.
2. Commutator subgroup is fully characteristic

Proof

For the field with two elements

In this case, every element of $GL_n(k)$ has determinant one, so $SL_n(k) = GL_n(k)$. Since every group is fully characteristic as a subgroup of itself, $SL_n(k)$ is a fully characteristic subgroup of $GL_n(k)$.

For other cases

In all other cases, the result follows from facts (1) and (2).