Bourbaki

Proposition 2 (Maschke)

Tag 0AU0, permanent. Cite it as [Bourbaki, Tag 0AU0].

Lie Groups and Lie Algebras, chapter V, Appendix 0. Complements on linear representations

Let A be a ring with unit element, E a left A-module, F a direct summand of E, G a group of finite order q, and ρ\rho a linear representation of G on E. Assume that q.1 is invertible in A and that F is stable under G. Then there exists a complement of F in E that is stable under G.

Let pp be a projection of EE onto FF. For all x∈Ex \in E, put

f(x)=q−1∑s∈Gρ(s)−1p(ρ(s)x). f(x) = q^{-1} \sum_{s \in G} \rho(s)^{-1} p(\rho(s)x).
We have f(x)∈Ff(x) \in F and f(y)=yf(y) = y for all y∈Fy \in F, so ff is a projection of EE onto FF. On the other hand, if t∈Gt \in G,
ρ(t)f(x)=q−1∑s∈Gρ(st−1)−1p(ρ(s)x)=q−1∑s∈Gρ(s)−1p(ρ(st)x)=f(ρ(t)x). \begin{align*} \rho(t)f(x) &= q^{-1} \sum_{s \in G} \rho(st^{-1})^{-1} p(\rho(s)x) \\ &= q^{-1} \sum_{s \in G} \rho(s)^{-1} p(\rho(st)x) \\ &= f(\rho(t)x). \end{align*}
Thus ff commutes with ρ(G)\rho(G), so ker⁡(f)\ker(f) is a complement of FF stable under GG.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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