Conjugacy for $p$-adic matrices of finite order

Question

$\DeclareMathOperator\GL{GL}$Say $p$ is an odd prime, and take two matrices $A,B\in \GL_n({\mathbb Z}_p)$ of finite order $m$. Is it true that they are conjugate in $\GL_n({\mathbb Z}_p)$ if and only if their reductions mod $p$ are conjugate in $\GL_n({\mathbb F}_p)$?

Edit: Thank you all for the answers and a very insightful discussion! As the answer is NO, it is important for me to know whether conjugacy in $\GL_n({\mathbb F}_p)$ at least implies conjugacy in $\GL_n({\mathbb Q}_p)$. I cannot see this from the answers, so I think I will ask this as a separate question.

Answer 1

I think I finally have a correct answer for arbitrary $p$.

As F. Ladisch notes, $G=C_{p^3}$ has only finitely many indecomposable modular representations. For the following argument, I will not only need infinitely many integral representations, but I will need them to occur at reasonably regular intervals, as the rank goes up. Luckily, they do. In "Representations of Cyclic Groups in Rings of Integers, II", Heller and Reiner exhibit indecomposable $\mathbb{Z}_p[G]$-modules of rank $kp^3$ for arbitrary $k\in \mathbb{N}$.

Now, let us count the number of arbitrary $\mathbb{F}_p[G]$-modules, respectively of arbitrary $\mathbb{Z}_p[G]$-modules, of rank $np^3$. The former correspond to partitions of $np^3$ into summands that come from a fixed finite set of integers (possibly some repetitions), and their number is easily seen to be polynomial in $n$ (e.g. if the finite set is $A$, then a very crude upper bound on the number of partitions is $\prod_{i\in A} np^3/i$).

On the other hand, even if we consider only direct sums of the above mentioned modules of rank $kp^3$, $k\in \mathbb{N}$, we get a number that is essentially the partition number of $n$, which is exponential in $\sqrt{n}$. So for sufficiently large $n$, we get much more $\mathbb{Z}_p[G]$-modules of rank $np^3$ than $\mathbb{F}_p[G]$-modules. This shows the even stronger claim that arbitrarily many non-conjugate matrices over $\mathbb{Z}_p$ can reduce to conjugate $\mathbb{F}_p$-matrices.

It is also worth remarking that for $G=C_p$, it is true that two $\mathbb{Z}_p[G]$-modules are isomorphic if and only if their reductions are. That's because the only indecomposable $\mathbb{Z}_p[G]$-modules are the trivial module, the regular module and the $p-1$-dimensional augmentation ideal in the regular module (at the moment, I can't find a good reference, so I might edit it in later). For $G=C_{p^2}$, one could also go through the finite classification (Reiner - Integral Representations of Cyclic Groups of Order $p^2$) and see whether this still holds.

Answer 2

Here is an infinite collection of "cheating" counterexamples for $p=2$. Let $G$ be a cyclic group of order $2^n ,$ generated by $g$, say. Let $M$ be the augmentation ideal of the group ring $\mathbb{Z}_{2}G,$ so that $M = \sum_{x \in G} \alpha_x x,$ where $\sum_{x \in G}\alpha_{x} = 0$. Regard $M$ as a (say, right) $\mathbb{Z}_{2}G$-module (of rank $2^{n}-1$, for example with $\mathbb{Z}_2$-basis $\{ x - 1_G: 1 \neq x \in G \}$). Note that the minimum polynomial of $g$ on $M$ is $\frac{t^{2^n}-1}{t-1}$, and note also that $g$ acts with determinant $-1$ on $M.$ Now let $V$ be a rank 1 $\mathbb{Z}_2G$-module on which $g$ acts as $-1$. Then $M \otimes V$ and $M$ have isomorphic reductions (mod 2), but are not isomorphic as $\mathbb{Z}_2G$-modules (since $g$ acts with determinant $1$ on the first, and determinant $-1$ on the second).

Answer 3

$\DeclareMathOperator\GL{GL}$I don't have a counter-example at the moment but the result seems too strong to be true. Here are some partial results instead.

Let $f\in \mathbb{Z}_p[x]$ be so that its reduction $\pmod{p}$ doesn't have repeated roots. Two matrices $A,B\in \GL_n(\mathbb{Z}_p)$, satisfying $f(A)=f(B)=0$, will be conjugate in $\mathbb{Z}_p$ provided that they are conjugate over $\mathbb{F}_p$. This is theorem 2 in Certain matrix equations over rings of integers by R.W. Davis.

If the orders of two matrices $A,B\in \GL_n(\mathbb Z/p^n\mathbb Z)$ are coprime to $p$, then they are similar in $\GL_n(\mathbb Z/p^n\mathbb Z)$ if and only if their reductions are similar in $\GL_n(\mathbb F_p)$. This is proved in the article "On the conjugacy of matrices over a ring of residues" by D.A. Suprunenko (MR0172886).

Answer 4

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis–Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been pointed out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis–Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k\lvert G\rvert$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k\lvert G\rvert$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k $, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2\lvert G\rvert$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$. (Added later: This is elementary. Remember that $1+ p M_n(\mathbb{Z}_p) \subseteq \operatorname{GL}_n(\mathbb{Z}_p)$, so units of $M_n(\mathbb{F}_p)$ lift to units of $M_n(\mathbb{Z}_p)$. After replacing $B$ with a conjugate, we may assume that $A\equiv B \mod p$. Then
$$ U:= \frac{1}{m} \sum_{k=0}^{m-1} A^{-k} B^k \equiv \frac{1}{m} \sum_{k=0}^{m-1} I \equiv I \mod p $$ which implies that $U$ is invertible. One computes that $AU=UB$, so it follows $A^U = B$.)

Answer 5

Here is a trial proof for the question over $Q_p$.

Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$).

Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers).

Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$ for $(u,p)=1$. Further note, if $\bar\Phi_u=\prod \bar g$ then $\bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$, and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split, in other words this is the minimal polynomial. This follows since the reduction (mod $p$) of the companion matrix of $f$ is itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$.

So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$.

From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. The general Jordan form classifies the conjugacy type over a field, as is $Q_p$.

Note that, $\Phi_3\Phi_6$ and $\Phi_6^2$ give 4x4 matrices with order 6, failing for $p=2$.

Answer 6

$\DeclareMathOperator\GL{GL}$Here is a proof in the case where $p \nmid m$ that doesn't involve modular representations, though it does involve a result of Gelfand and Kazhdan whose proof I can't remember, so I have no idea whether it's easy or hard.

A special case of their result is that two elements of $\GL_n(\mathbb{Z}_p)$ are conjugate in that group if and only if they are conjugate in $\GL_n(\mathbb{Q}_p)$. Moreover, two elements of $\GL_n(\mathbb{Q}_p)$ are conjugate in that group if and only if they are conjugate in $\GL_n(E)$ for every extension $E/\mathbb{Q}_p$. Thus, if we assume that the reductions $\bar A$ and $\bar B$ of $A$ and $B$ are conjugate in $\GL_n(\mathbb{F}_p)$, it will be enough to show that $A$ and $B$ are conjugate in $\GL_n(E)$ for some $E$.

Let $E/\mathbb{Q}_p$ be an extension containing all $m$th roots of unity. These roots of unity, and thus the eigenvalues of $A$ and $B$, are completely determined by their images in the residue field of $E$, which are the eigenvalues of $\bar A$ and $\bar B$. That is, $A$ and $B$ have the same multi-set of eigenvalues if and only if $\bar A$ and $\bar B$ do.

I believe that one could construct another proof from the result about $\GL_n(\mathbb{Z}/p^n\mathbb{Z})$ mentioned in Gjergji's answer [at least if he really meant $\GL_n(\mathbb{Z}_p/p^k\mathbb{Z})$], via some limiting process.