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 $G l_{n} (k)$ 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 $G L_{n} (k)$ of invertible $n \times n$ matrices, the subgroup $D_{n} (k)$ of invertible diagonal matrices.

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

Proof: Suppose $A \in G L_{n} (k)$ 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 $(i j)^{t h}$ entry. Consider a diagonal matrix $B$ with different entries in the $(i i)$ and $(j j)$ places (for this, we need the field to have more than two elements). Then, $B A$ and $A B$ have different values in the $(i j)^{t h}$ position, hence $A B \neq B A$ . Hence, $A$ does not commute with every element of $D_{n} (k)$ , leading to the required contradiction.