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> 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. 
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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!



Centralizers of elements in general linear group over Z mod prime power
Let $G\_r=\mathrm{GL}\_n(\mathbb{Z}/p^r)$. For $x\in G\_r$ the centraliser $C\_{G\_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, *Regular elements and regular characters of* $\mathrm{GL}\_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.


To add some details, note that Hill's Theorem 3.6 holds when the residue field is $\overline{\mathbb{F}}\_p$, but I think his proof goes through for any algebraically closed residue field $\overline{k}$. If $k$ is any field and $C\_{\mathrm{GL}\_n(k)}(A)$ is abelian for $A\in \mathrm{GL}\_n(k)$, then using for example the rational canonical form one can see that $A$ must be $\mathrm{GL}\_n(k)$-conjugate to a companion matrix, and so $k^n$ is a cyclic $k[A]$-module. Extending scalars to $\overline{k}$ we get that $A$ is regular as an element of $\mathrm{GL}\_n(\overline{k})$.


