Let $p$ be a prime and $n, k$ positive integers. The group $GL_n(\mathbb{Z}/p^k \mathbb{Z})$ contains the principal congruence subgroup $GL_n(\mathbb{Z}/p^k \mathbb{Z})_1 = \{I_n + p M : M \in M_n( \mathbb{Z}/p^k \mathbb{Z} ) \}$, which is the kernel of the reduction modulo $p$.
Every finite group can be embedded into $GL_n(\mathbb{Z}/p^k \mathbb{Z})$ via permutation matrices, for $n$ large enough. For $GL_n(\mathbb{Z}/p^k \mathbb{Z})_1$ we need to restrict to finite $p$-groups, and moreover we need to be able to take larger $k$, for instance to avoid having a bound on the solvable length. But besides that, can we embed every group?
That is, given $P$ a finite $p$-group, $k, n$ large enough, does $P$ embed in $GL_n(\mathbb{Z}/p^k \mathbb{Z})_1$?
As an example, we can embed every abelian $p$-group: take $k = 2m$, then the subgroup of matrices congruent to the identity modulo $p^m$ is isomorphic to the additive group of $M_n(\mathbb{Z}/p^m \mathbb{Z})$.