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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.

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!

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  • $\begingroup$ You want a reference for the case $e=1$ and $n=2$? This follows easily from the canonical rational form and also answers your question for general $n$ and $e=1$. $\endgroup$
    – Marc Palm
    Commented Oct 7, 2013 at 10:46
  • $\begingroup$ For general $e$, it might be useful to fix $\gamma \in GL_n(\mathbb{Z}_\ell)$ and argue inductively, e.g. via maps $GL_n(\mathbb{Z}/\ell^d) \rightarrow GL_n(\mathbb{Z}/\ell^k)$ and how the results vary as does $e$, e.g., the centralizer may become non-commutative as $e$ grows. $\endgroup$
    – Marc Palm
    Commented Oct 7, 2013 at 10:49
  • $\begingroup$ This might be usefull: arxiv.org/abs/0807.4684. It discusses the classificiation for irreducible representations (still open for $n>2$), which are in some sense dual to the conjugacy classes and centralizer. From that perspective, I would be suprised if the conjugacy classes have been classified for $GL_n(\mathbb{Z}/\ell^d)$ let alone their centralizers. $\endgroup$
    – Marc Palm
    Commented Oct 7, 2013 at 10:51

1 Answer 1

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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 $\mathrm{GL}_n(\overline{k})$.

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    $\begingroup$ Dear Alexander, thanks for your answer and for deducing your nice corollary out of Theorem 3.6 of the given reference. If I understand it right, then I can even go one step further and say: Take $x \in G_r$ and denote by $\overline x$ the reduction of $x$ modulo $p$. Then $C_{G_r}(x)$ is abelian if and only if $C_{GL_n(\mathbb{F}_p)}(\overline x)$ is abelian. $\endgroup$
    – Stefan Keil
    Commented Oct 7, 2013 at 16:42

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