Representation pullbackability theorem

This fact is related to: representation theory
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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 ${\tilde {\rho }}$.