Diagonal subgroup is self-centralizing in general linear group

This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a group being self-centralizing. In other words, the centralizer of the subgroup in the group is contained in the subgroup
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Statement

Suppose $k$ is a field with more than two elements. Then, consider the general linear group $Gl_{n}(k)$ of invertible $n\times n$ matrices over $k$. The subgroup $D_{n}(k)$ of invertible diagonal $n\times n$ matrices is self-centralizing.

(If $k$ has only two elements, the diagonal subgroup is trivial, and so clearly is not self-centralizing for $n\geq 2$).

Proof

Given: A field $k$ with more than two elements, the group $GL_{n}(k)$ of invertible $n\times n$ matrices, the subgroup $D_{n}(k)$ of invertible diagonal matrices.

To prove: $D_{n}(k)$ is self-centralizing in $GL_{n}(k)$.

Proof: Suppose $A\in GL_{n}(k)$ has the property that $A$ commutes with every element of $D_{n}(k)$. We want to show that $A\in D_{n}(k)$.

Suppose not. Then, there exists a nonzero off-diagonal entry of $A$, say the $(ij)^{th}$ entry. Consider a diagonal matrix $B$ with different entries in the $(ii)$ and $(jj)$ places (for this, we need the field to have more than two elements). Then, $BA$ and $AB$ have different values in the $(ij)^{th}$ position, hence $AB\neq BA$. Hence, $A$ does not commute with every element of $D_{n}(k)$, leading to the required contradiction.