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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.

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I think that the name of Alfredo Jones is relevant here. Try the 1962 Curtis and Reiner, and their later "Methods of Representation Theory" tome. – Geoff Robinson Aug 5 '11 at 13:01
Afterthought: Since the answer to your first question is "No", it may also be useful to know that conjugacy mod $p^r$ implies conjugacy in $GL_n(\mathbb{Z}_p)$, when $r$ is big enough, i. e. such that $p^r$ doesn't divide $m$. This follows from the treatment of Maranda's theorem in Curtis-Reiner, Methods. – Frieder Ladisch Aug 10 '11 at 19:36
up vote 13 down vote accepted

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 and see whether this still holds.

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Very nice. The first paper by Heller and Reiner (Representations of Cyclic Groups in Rings of Integers, I) explains why there are only 3 indecomposable $\mathbf{Z}_p[C_p]$-modules in Theorem 2.6, and also treats the case $G = C_{p^2}$. The key idea is that any $\mathbb{Z}_p[C_p]$-module $M$ is an extension of a $B$-module $M/N$ by an $A$-module $N$; here $B = \mathbb{Z}_p[\zeta_p]$ and $A = \mathbb{Z}_p$ are both discrete valuation rings, so $M/N$ is a free $B$-module and $N$ is a free $A$-module whenever $M$ is $p$-torsion-free. Heller and Reiner explain how to handle the $Ext^1$ group. – user91132 Aug 6 '11 at 12:18
Thanks for the complement and for the reference, Kostia! – Alex B. Aug 6 '11 at 12:23
Really good. Dade shows more generally that $\mathbb{Z}_pG$ has indecomposable lattices of rank $kp^3$ whenever $\mathbb{Q}_pG$ has at least $4$ different irreducible representations. (In fact, the result is even more general.) Dade's paper is in the same issue of the Annals as the Heller-Reiner II, but is also in the first volume of Curtis-Reiner (Methods...). – Frieder Ladisch Aug 6 '11 at 13:07
I am glad that is finally resolved. – Geoff Robinson Aug 6 '11 at 13:27
Note that this method also answers the question about $\mathbb{F}_p$ and $\mathbb{Q}_p$ conjugacy. There are only four indecomposable representations of $\mathbb{Q}_p[\mathbb{Z}/p^3]$, so the number of isomorphism classes of $\mathbb{Q}_p$ representations of dimension $n$ is $O(n^4)$ and the number of joint pairs $(\mathbb{F}_p \mathrm{-rep}, \mathbb{Q}_p \mathrm{-rep})$ is polynomial in $n$. – David Speyer Aug 8 '11 at 16:18

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).

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Great idea. Even more simply, let $A$ be any matrix of odd dimension and order $2^r$, and take $B=-A$. – Frieder Ladisch Aug 5 '11 at 13:44
How is that cheating? You've solved the problem (for $p=2$). – David Speyer Aug 5 '11 at 13:59
@David: Well it isn't really cheating, but relies on having an extra $p$-th root of unity in the unramified case when $p = 2$. – Geoff Robinson Aug 5 '11 at 14:04

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 it's 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).

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The case $m = p$ and $n = p-1$ is particularly interesting. – user91132 Aug 5 '11 at 11:57
I had the opposite thought. The forcing of $A,B$ to have finite order in $GL_n(Z_p)$ is so strong, that it eliminates any mischief on the periphery of your above theorems. Yet I also don't have a proof, or counterexample. The essential line of thought, which I made was to conjugate to blocks of a general Jordan form (companion form of a cyclotomic polynomial), and the finite order condition restricts the glueing elements to be zero. Then if this simplification is ok, it seems a tricky question about whether two cyclotomic polynomials can be congruent mod an odd prime. – Junkie Aug 5 '11 at 12:24
@Junkie: really? I would have expected problems if $p | m$ as these results suggest (by analogy with Hensel's lemma, for example). – Qiaochu Yuan Aug 5 '11 at 14:27
You are right, Qiaochu Yuan, the best answer above noted in essence that $\Phi_{p^3}\equiv\Phi_{p^2}^p$ modulo $p$, and I did not think this to be possible, but I think I should have suspected this, for how $p$th powers work modulo $p$. – Junkie Aug 6 '11 at 4:00
However, the general Jordan form for $\Phi_{p^3}$ is $p$ copies of the $\Phi_{p^2}$ matrix glued by 1's, while the direct sum from $\Phi_{p^2}^p$ does not have these extra 1's. I think it is true that the general Jordan forms of companion matrices of factors of cyclotomic polynomials, when taken modulo $p$ are enough distinct that the mod $p$ decomposition determines that in $Z_p$. – Junkie Aug 6 '11 at 9:19

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 poited 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 $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ 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 $2|G|$.
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 $1+ p M_n(\mathbb{Z}_p) \subseteq GL_n(\mathbb{Z}_p)$, 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 $A^U = B$.)

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Could you briefly explain why you can always lift the modular matrices to integral matrices of order dividing $p^3$? – Alex B. Aug 5 '11 at 16:23
@F.Ladisch: I am confused. Isn't it true that if $p>n$ then a pro-$p$ Sylow of $GL_n(\mathbb Z_p)$ does not contain an element of finite order. If I recall this correctly this has been remarked by Lazard in his IHES paper on p-adic analytic groups. So if $p> n$ then all the elements of finite order in $GL_n(\mathbb Z_p)$ have order prime to $p$. So your answer should say "Yes, for almost all primes". – vytas Aug 5 '11 at 17:43
@Geoff Robinson. I am saying that if $p-1> n$ then $\GL_n(\mathbb Z_p)$ does not contain an element of order $p$. Given that, and the last sentence in F.Ladisch's answer, we obtain: "Yes, if $p-1>n$". Ah! I have uncofused myself, Ladisch is saying for every prime $p$ there exists an $n$, such that the statement is false, I am saying for a fixed $n$ the statement is true for almost all primes $p$. – vytas Aug 5 '11 at 18:12
Put another way I think, $\Psi_{p^3}\equiv\Psi_{p^2}^p$ modulo $p$ for the cyclotomic polynomials $\Psi$ and from this the desired matrices can appear with companions. EDIT: The companions $A$ and $B^{\oplus p}$ do not have the order ($p^3$ resp. $p^2$), but summing each to $A$ will. However, I am still unsure it all works, for the final matrices are not conjugate in $F_p$ when I calculate. – Junkie Aug 6 '11 at 4:22
With my above comment, it seems that the companion matrix of $\Psi_{p^3}$ when reduced mod $p$ has a different (Jordan) form than $p$ copies of $\Psi_{p^2}$ when resp. reduced, so this direction in cyclotomic polys is murky. I am back to looking at Alex Bartel's comment, as to why the lifting step should occur. – Junkie Aug 6 '11 at 4:54

Here is a proof in the case where $p \not |m$ 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.

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What is the result of Gelfand and Kazhdan that you refer to? I suppose that the corollary that you mention ("A special case of their result...") already uses the fact that the two elements have order coprime to $p$. – Tim Dokchitser Aug 7 '11 at 9:52
The result can be found in Gelfand-Kazhdan, "Representations of the group GL(n,K) where K is a local field. Lie groups and their representations (Proc. Summer School, Bolyai Janos Math. Soc., Budapest, 1971)", published in 1975. Sorry, but I don't have my files handy, and I'm suddenly unsure if the result is for elements of order co-prime to p, or all elements. But either way, it says that two elements of a maximal compact subgroup are conjugate in the subgroup if and only if they are conjugate in GL(n,K). – Jeff Adler Aug 7 '11 at 17:52
Addendum: The Gelfand-Kazhdan result is special for GL(n). It certainly doesn't work for reductive p-adic groups in general. – Jeff Adler Aug 7 '11 at 17:54
It should probably be for order coprime to $p$; e.g. in $GL_2(Z_2)$ the reflection in the x-axis and the reflection in the line x=y (A=[[1,0],[0,-1]] and B=[[0,1],[1,0]]) are not conjugate, but they are conjugate over $Q_2$. – Tim Dokchitser Aug 8 '11 at 8:43
When $p \nmid m$, the result is rather elementary, see the addendum to my answer. The argument (averaging) is quite common in group representation theory, but no knowledge about representations is needed to follow the argument, I think. – Frieder Ladisch Aug 8 '11 at 10:44

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 $p=2$.

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This sounds like the right approach, but I am worried about the "Furthernote,..." step. Why is this true? I can only see that in every $Q_p$-conjugacy class there is a matrix for which the claim holds. E.g, in particular, this step implies immediately that a matrix of order $m>1$ can't reduce to identity. This is indeed true (when $p \ne 2$), but I think this is a non-trivial statement. Or am I wrong? Also, "This classifies it over a field" is worrisome, because the statement is false over $Q_p[\mu_p]$. – Tim Dokchitser Aug 8 '11 at 8:39
For the last difficulty, do you need to change the $\phi$-function definition over $Q_p(\mu_p)$, so that the factor of $(p-1)$ disappears into the units? In other words, there is additional splitting over $Q_p(\mu_p)$ viz. $Q_p$, but do you start at $F_p(\mu_p)$ now also? I would say, my statement "this classifies it" corresponds to the Jordan form part of the argument, not the rest of it where the reduction is done. That part uses that $Q_p$ and $F_p$ splitting are the same, up to ramification, but $Q_p(\mu_p)$ can have more. – Junkie Aug 8 '11 at 10:35

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