TITLE: Torsion in GL_n(Z)
QUESTION [18 upvotes]: 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 $g \in \text{ker}(\phi_p)$.

REPLY [2 votes]: 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 $y$.