Let \(G\) be a finite \(p\)-group and let \(V\) be a nonzero finite-dimensional vector space over a finite field \(k\) of characteristic \(p\). What can be said about a representation \[ G\to GL(V)? \]
Solution
Every such representation has a nonzero fixed vector: \[ V^G\ne0. \]
Indeed, let \(G\) act on the finite set underlying \(V\). Every orbit has size a power of \(p\). Hence all nontrivial orbits have cardinality divisible by \(p\). Therefore \[ |V|\equiv |V^G|\pmod p. \] Since \[ |V|=|k|^{\dim V} \] is divisible by \(p\), it follows that \[ |V^G|\equiv0\pmod p. \] The zero vector is fixed, so \(|V^G|\ge p\) and there is a nonzero fixed vector.
Consequently every irreducible \(k\)-representation of \(G\) is trivial. If \(V\) is irreducible, then the nonzero fixed subspace \(V^G\) is a nonzero \(G\)-stable subspace, so \[ V^G=V. \] Thus \(G\) acts trivially. Irreducibility then forces \(\dim_kV=1\).
More generally, every finite-dimensional representation admits a filtration whose successive quotients are trivial one-dimensional representations: choose a nonzero fixed vector, quotient by its span, and argue inductively.