A question about $GL_n(\mathbb{Z})$ and $GL_n(\mathbb{F}_p)$ Let $p\ge 3$ be a prime number. $G$ is a subgroup of $GL_n(\mathbb{Z})$ and $|G|<\infty$. Let $\sigma: GL_n(\mathbb{Z})\to GL_n(\mathbb{F}_p)$ be the natural map. Prove that $\sigma|_G$ is injective.
Suppose $\exists A,B \in G,A\ne B$, s.t. $\sigma(A)=\sigma(B)$ i.e. $A=B \pmod{ p}$. Since $|G|<\infty$, $\exists m,k\in\mathbb{Z}^+$, $A^m=B^k=I$. In linear algebra we know $A=C\ {\rm diag}(\zeta_1,\dots,\zeta_n)\ C^{-1}$ where $\zeta_i^m=1,C\in GL_n(\mathbb{C})$. But I don't know if $m=k$. Taking trace and norm can't solve this problem. I think $A=B \pmod{p} $ is not easy to use.
Any ideas?
 A: Let $A \in \mathrm{Ker}(\sigma_{|G})$. Then $A \equiv I_n \ [\mathrm{mod}\ p]$, so there exists $N \in \mathcal{M}_n(\mathbb{Z})$ such that $A = I_n + pN$.
But by Lagrange theorem, $A^{|G|}=I_n$, so $(I_n + pN)^{|G|}=I_n$. So the polynomial $(1+pX)^{|G|}-1$ is a vanishing polynomial for $N$, and since it is splitted and has simple roots over $\mathbb{C}$, then $N$ is diagonalizable, and the eigenvalues of $N$ are among its roots, which are the
$$\frac{e^{2ik\pi/|G|}-1}{p}, \quad \quad k=1, ..., |G|$$
These eigenvalues have modulus $<1$ (because $p\geq 3$), so $\lim_{k \rightarrow +\infty} N^k = 0$. But $N \in \mathcal{M}_n(\mathbb{Z})$, so the sequence $(N^k)_{k \geq 1}$ is eventually constant equal to $0$, so $N$ is nilpotent.
So $N$ being diagonalizable and nilpotent, then $N=0$, so $A=I_n$ and you are done.
A: To show injectivity, it's enough to show that if $A\in G$ and $\sigma(A) = 1$, then $A = 1$.
Suppose that $\sigma(A) = 1$. Then we can write $A = I + pB$ where $B\in \mathrm{M}_2(\mathbb Z)$.
Let $d$ be the greatest common divisor of the entries of $B$, so that $B = dC$, and the greatest common divisor of the entries of $C$ is $1$.
By assumption, $A^q = I$ for some $m\in \mathbb Z$. Replacing $A$ by some power, we can assume that $q$ is prime.  So
$$I=A^q = (I + pdC)^q= \sum_{i = 0}^q{q\choose i}p^id^iC^i$$
Rearranging, we have
$$qdpC = -\sum_{i=2}^q{q\choose i}(pd)^iC^i$$
so
$$qC = -pd\sum_{i=2}^q{q\choose i}(pd)^{i-2}C^i.$$
Since, by assumption, $C$ has no common factors, we have $q = p$ and $d=1$.
So
$$C = -\sum_{i = 2}^p{p\choose i}p^{i-2}C^i,$$
which gives a contradiction whenever $p\ge 3$, as $p\mid{p\choose i}$, so the RHS is divisible by $p$, but the LHS is not.
