Torsion in GL_n(Z) Fix some $n \geq 3$.  It's hopeless to classify the torsion elements in $\text{GL}_n(\mathbb{Z})$, but I have a couple of less ambitious questions.  It's well-known that for any odd prime $p$, the map $\phi_p : \text{GL}_n(\mathbb{Z}) \rightarrow \text{GL}_n(\mathbb{Z}/p)$ is ``injective on the torsion''.  In other words, if $F$ is a finite subgroup of $\text{GL}_n(\mathbb{Z})$, then $\phi_p|_F$ is injective.
Fixing an odd prime $p$, I have two questions about the map $\phi_p$.


*

*Are there any interesting restrictions on elements $y \in \text{GL}_n(\mathbb{Z}/p)$ such that there exists some finite-order $x \in \text{GL}_n(\mathbb{Z})$ with $\phi_p(x) = y$?

*Do there exist any non-conjugate finite order elements $x,x' \in \text{GL}_n(\mathbb{Z})$ such that $\phi_p(x) = \phi_p(x')$?  The hypothesis that $x$ and $x'$ are non-conjugate is to rule out the silly example where $x' = g x g^{-1}$ with $g \in \text{ker}(\phi_p)$.
 A: A finite order element  $A\in GL(n,\mathbb{Z})$ has eigenvalues which are roots of unity, and its characteristic polynomial $det(\lambda I-A)$ is a product of cyclotomic polynomials. 
Thus, it is a necessary condition that the $(\mod p)$ reduction of $A$ have characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials. 
Conversely, suppose that $A_p\in GL(n,\mathbb{Z}/p)$ has a characteristic polynomial which is a $(\mod p)$ reduction of a product of cyclotomic polynomials. Moreover, assume that $A_p$ is
conjugate in $GL(n,\mathbb{Z}/p)$ to the companion matrix of its characteristic polynomial (one may check this algorithmically by putting the matrix in rational canonical form over $\mathbb{Z}/p$). Then clearly it is the $(\mod p)$ reduction of a finite order matrix in $GL(n,\mathbb{Z})$. I think this criterion is necessary and sufficient, but I haven't checked it. 
A: Let $m$ be the order of $y$.
If $p-1>n$ then a pro-$p$ Sylow of $\mathrm{SL}_n(\mathbb Z_p)$ is torsion free, see (3.2.7.5) on p.101 of http://www.numdam.org/item?id=ASENS_1954_3_71_2_101_0. Hence, if $p-1>n$ and $p$
divides $m$, then $y$ cannot be lifted.
If $y$ can be lifted to a torsion element $x$, then the  order of $x$ is $m$, by "injection of
torsion", and hence the characteristic polynomial $\chi$ of $y$ can be lifted to a monic polynomial $\tilde{\chi}$ of degree $n$ in $\mathbb Z[X]$, such that all irreducible factors 
of $\tilde{\chi}$ divide $X^m-1$. (this is also pointed out by Agol above).
If $p$ does not divide $m$ then the converse is also true. 
Proof. Suppose $\tilde{\chi}=f_1^{n_1}\ldots f_k^{n_k}$, with $f_i$ irreducible and distinct.
Let $M$ be a block diagonal matrix, with the companion matrix of $f_i$ coming up $n_i$ times.
The char poly of $M$ is $\tilde{\chi}$, the minimal poly of M is $f_1\ldots f_k$, which divides $X^m-1$. Hence $M^m=1$ and thus $M\in\mathrm{SL}_n(\mathbb Z)$. 
The image of $M$ in $\mathrm{SL}_2(\mathbb{F}_p)$ is conjugate to $y$, since 
they have the same char poly and are both semisimple, as $p$ does not divide $m$.
Hence one can lift $y$.
A: Here is a loose collection of partial answers, most of which address a simpler question than that Andy actually asked. I am particularly focusing on Andy's question in the comments "can we classify all liftable elements of ${\rm GL}_n(\mathbb{Z}/p\mathbb{Z})$ of order $k$ for some fixed $k$?". Please take all this with a grain of salt, because I am improvising as I go and it's already late at my end.
The first remark is that I don't think that restricting the order of the element makes the problem any easier. That's because, suppose that you have an element $X$ of order $k$ that lifts from ${\rm GL}_n(\mathbb{Z}/p\mathbb{Z})$ to ${\rm GL}_n(\mathbb{Z})$. Then, for any invertible $A\in {\rm GL}_n(\mathbb{Z}/p\mathbb{Z})$, to find out whether $AXA^{-1}$ lifts, you will almost certainly have to determine whether $A$ lifts. But $A$ can be of arbitrary order.
To fix this, I will instead consider the question: for a given $k$, what modular representations of the cyclic group of order $k$ can be lifted to integral ones. This also bypasses another issue: namely that the order of the lift could potentially be higher than that of the original element.
Edit: please ignore that last sentence. By injection of torsion, this can't happen, as pointed out by vytas.
One fact that I will implicitly use all the time is that if two integral representations give rise to the same rational representation, then their reductions have the same semi-simplifications.
Case 1: Suppose that $k$ is co-prime to $p$ and that the order of $p$ in $(\mathbb{Z}/k\mathbb{Z})^\times$ is $\varphi(k)$, in other words that the extension of $\mathbb{F}_p$ obtained by adjoining the $k$-th roots of unity is as big as possible. This ensures that it's "no easier" for a representation of $C_k$ to be defined over $\mathbb{F}_p$, than over $\mathbb{Q}$ (equivalently than over $\mathbb{Z}$). Then, by construction, there is a bijection between indecomposable mod $p$ representations and indecomposable integral representations of $C_k$, thus any representation lifts. But as I remarked at the beginning, that probably doesn't mean that any element of order $k$ lifts. (Or does it? Can someone fix this issue?)
Case 2: Let's just assume that $k$ is co-prime to $p$. The previous case should give a hint for this more general one: there is a bijection between representations of $C_k$ over a sufficiently large extension of $\mathbb{F}_p$ and representations over a sufficiently large extension of $\mathbb{Q}$. For example if $k$ is prime, then the irreducible rational representations are the trivial and a $p-1$-dimensional. Therefore, a representation can be lifted if and only if the numbers of copies of all the non-trivial representations of $C_k$ over $\mathbb{F}_p(\zeta_k)$ in this representation are equal. Note that talking about liftable representations, rather than matrices allows me to ignore the very subtle issue of non-isomorphic integral representations sitting in the rational one - they all give the same reduction modulo $p$. If $k$ is composite, then a similar analysis can be carried out.
Case 3: Suppose that $k=p$ or $k=p^2$. There is one indecomposable modular representation of dimension $n$ for each $n$ between 1 and $p$. Their semi-simplifications are all sums of trivial representations. Moreover, if I remember correctly, the reduction of any integral representation sitting inside the irreducible rational one of dimension $p-1$ remains indecomposable over $\mathbb{F}_p$. Thus, a representation is liftable, if and only if it is a direct sum of indecomposable representations of dimensions 1 and $p-1$. I believe that with some more work, the case $k=p^2$ can be worked out in a similar vein.
Case 4: $k$ is divisible by $p^3$. Here, all hell breaks loose, because now, there are infinitely many isomorphism classes of indecomposable integral representations and they can have arbitrarily large ranks. No one knows how to classify them, so it's hopeless to try listing their reductions. However, since you are fixing the dimension $n$, there is some good news:
Jordan-Zassenhaus Theorem: there exist only finitely many isomorphism classes of integral representations giving rise to any given rational representation.
If I remember correctly, the proof constructs these for you (it breaks up into two parts: first showing that there are only finitely many isomorphism classes in each genus (i.e. integral representations that are locally isomorphic everywhere) and then showing that there are finitely many genera in any given rational representation). So for any given $n$ and $k$, you can follow through the proof to get a classification of integral representations and therefore of their reductions. As far as I know, there is no uniform description of the answer, you just have to compute it in any given case.
I hope, I didn't write completely off topic, seeing as I didn't really address the actual question.
For most of the facts about integral representations together with proper references, see Reiner's excellent survey article. The modular stuff can be found e.g. in Benson's book, or, in those cases, where I messed up, not at all.
