Representation pullbackability theorem

Statement
Let $$G$$ be a finite group and $$k$$ a finite field whose characteristic does not divide the order of $$G$$. Let $$A$$ be a local ring with residue field $$k$$. Then, any representation of $$G$$ over $$k$$ pulls back to a representation of $$G$$ over $$A$$.

In other words, given any map:

$$\rho:G \to GL_n(k)$$

there exists a map:

$$\tilde{\rho}:G \to GL_n(A)$$

such that $$\tilde{\rho}$$, composed with the residue map modulo the maximal ideal, gives $$\rho$$.

Proof
$$k$$ has order $$q$$ where $$q=p^l$$, $$p$$ not dividing the order of $$G$$. Let $$\pi$$ be the map from $$GL_n(A)$$ to $$Gl_n(k)$$, that involves taking each matrix entry modulo the maximal ideal.

The kernel of $$\pi$$ is, as can easily be checked, the subgroup comprising matrices which are congruent entry-wise to the identity matrix. The cardinality of this kernel is thus $$p^{ln^2}$$.

Then, consider $$\pi^{-1}(\rho(G))$$. The map:

$$\pi: \pi^{-1}(\rho(G)) \to \rho(G)$$

has, as kernel, a $$p$$-group. Since $$p$$ does not divide the order of $$G$$, it does not divide the order of $$\rho(G)$$, os the kernel is a normal Sylow subgroup. We can thus apply Schur-Zassenhaus theorem to find a section for this map. Composing this section with $$\rho$$ gives us the pulled back representation $$\tilde{\rho}$$.