State and prove Maschke’s theorem. What can go wrong if you work over the real field? What can go wrong in characteristic \(p\)?
Solution
Maschke’s theorem. Let \(G\) be finite and let \(k\) be a field with \[ \operatorname{char}k\nmid |G|. \] Then every finite-dimensional \(k\)-representation of \(G\) is completely reducible.
Let \(W\subseteq V\) be \(G\)-stable and choose any linear projection \[ p_0:V\to W. \] Average it over the group: \[ p=\frac1{|G|}\sum_{g\in G}g\,p_0\,g^{-1}. \] The scalar \(|G|^{-1}\) exists by hypothesis. The map \(p\) is \(G\)-equivariant, has image \(W\), and restricts to the identity on \(W\). Hence \[ V=W\oplus\ker p, \] with \(\ker p\) also \(G\)-stable.
Over \(\mathbb R\), nothing fails in Maschke’s theorem: characteristic \(0\) still gives complete reducibility. What changes from the complex theory is the form of irreducibles and Schur’s lemma; for an irreducible real representation, the division algebra \(\operatorname{End}_G(V)\) can be \(\mathbb R\), \(\mathbb C\), or \(\mathbb H\).
If \(\operatorname{char}k=p\) divides \(|G|\), semisimplicity can fail. For \[ G=C_p=\langle g\rangle, \qquad k=\mathbb F_p, \] take \[ \rho(g)=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \] Because \((I+N)^p=I\) in characteristic \(p\) when \(N^2=0\), this is a representation of \(C_p\). The line spanned by \(e_1\) is invariant, but it has no invariant complementary line: the only eigenline of \(\rho(g)\) is \(\langle e_1\rangle\). Thus complete reducibility fails.