Diagonal subgroup is self-centralizing in general linear group

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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 \ge 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 \ne BA. Hence, A does not commute with every element of D_n(k), leading to the required contradiction.