Representation pullbackability theorem

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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}.