I would like to know for which elements $x$ in $G:=Gl_n(\mathbb{Z}/\ell^e\mathbb{Z})$ their centralizers $C_G(x):=\{ y \in G \mid xy=yx\}$ are abelian groups.

Here, $n$ is an integer $\geq 2$ and $\ell^e$ is a prime power. I am especially interested in the case $n=2$.

Of course, if $x$ is a multiple of the identity matrix, then $C_G(x)=G$, and thus is not abelian.

In case $e=1$, i.e. $\mathbb{Z}/\ell^e\mathbb{Z}$ is a field, then I have read a few times (but always without proof) that $C_G(x)$ is abelian, if $x$ is not a multiple of the identity matrix and $n=2$. To be precise, $C_G(x)$ will be either isomorphic to $\mathbb{F}_{\ell^2}^*$, $\mathbb{F}_\ell^* \times \mathbb{F}_\ell^*$, or $\mathbb{F}_\ell \times \mathbb{F}_\ell^*$.

So the question is what happens for arbitrary prime powers (and for arbitrary $n$)?

The question seems to me like a standard fact which should be contained in every text book about general linear groups. Can anybody give a good reference?

Thanks a lot!