Let $m,n,p\geq3$ be natural numbers and $G_1=\{A_1,A_2,...,A_p\}$ a multiplicative group of order $p$ with elements from $M_2(\mathbb{Z})$. For every $A_i$ from $G$ we attach $B_i$ such that if $A=(a_{ij}), i,j=1,n$, then $B=(a_{ij}) \pmod m$. Prove that the set $G_2=\{B_1,B_2,...,B_p\}$ is a multiplicative group isomorphic with $G_1$, where the multiplication is that of $M_n(\mathbb{Z}_m)$.
I think $f:G_1\rightarrow G_2$, $f(A_i)=B_i$ is an isomorphism and thus the problem can be easily solved, but I am unsure it's so basic. Any help to finish it?
This is not completely trivial. The map $f$ is certainly a homomorphism, but to show that it is an isomorphism, you need to show that its kernel is trivial.
This follows from the fact that the kernel of the reduction modulo $m$ map ${\rm SL}(n,{\mathbb Z}) \to {\rm SL}(n,{\mathbb Z}/m{\mathbb Z})$ is torsion-free for $m \ge 3$. You can find a proof of this in Lemma 2.6 here, for example.