When does the set of diagonal matrices over a finite prime field ($GL(n, F)$) form a normal subgroup? 
When does the set of diagonal matrices over a finite prime field ($GL(n, F)$) form a normal subgroup?

I know that the subgroup of diagonal matrices in the $Gl(n,F)$ is not closed under conjugation. However, does that happen over some finite field?
 A: It can only happen if $F$ is the field of two elements (or $n=1$).
If $n=1$ this is immediate. Assume $n\gt 1$.
If $|F|=2$, the only diagonal matrix in the general linear group is the identity, which is central.
If $F$ has more than two elements and $n\geq 2$, interpret the matrices as linear transformations on $F^n$; conjugation is equivalent to a change of basis. If the set of diagonal invertible matrices were normal, then given an invertible diagonalizable operator $T$ on $F^n$, every basis would consist of eigenvectors; that is, every vector would necessarily be an eigenvector.
But this cannot occur when $F$ has more than one nonzero element. Indeed, take the diagonal matrix with $\alpha$ in the $(1,1)$ coordinate, $\alpha\neq 1$, and $1$s elsewhere in the diagonal. Then the vector $(1,1,0,\ldots,0)$ is not an eigenvector, since it is mapped to $(\alpha,1,0,\ldots,0)$. So conjugating the matrix by the change-of-basis matrix that corresponds to going from the standard basis $\mathbf{e}_1,\ldots,\mathbf{e}_n$ to the basis $\mathbf{e}_1+\mathbf{e}_2,\mathbf{e}_2,\ldots,\mathbf{e}_n$ shows that the subgroup is not normal. 
Note that there is no assumption that $F$ is a prime field. The argument works for any field, finite or infinite, so long as it has a nonzero element different from $1$. 
