Any irreducible representation of a $p$-group over a field of characteristic $p$ is trivial.

Question

In general, we know that if $G$ is a finite group and $K$ is a field, then $K[G]$ (the group algebra) is semisimple whenever $\operatorname{char}(K)$ does not divide the order of $G$. However, this result does not hold when we have a $p$-group and a finite field of characteristic $p$.

How would I go about showing that an irreducible representation of a $p$-group $G$ over a field $K$ of characteristic $p$ must be the trivial representation?

It seems like this would perhaps entail some sort of application of Maschke's Theorem to get a contradiction. I know that $|G| = \sum_i \operatorname{dim}(V_i)^2$, where $V_i$ is an irreducible representation. But it seems like that does not help me much here.

Any thoughts as to how to approach this question?

Answer 1

Hint. Let $G$ act on the representation (minus $0$) and use an orbit counting argument to find a copy of the trivial representation.

Let $V$ be a nontrivial irreducible representation. Note that $V$ must be finite dimensional since $G$ is finite. Write $V^\circ = V\setminus\{0\}$ and let $G$ act on $V^\circ$. By orbit stabilizer, $\left| \mathcal{O}_x \right|$ divides $|G|$ for any $x\in V^\circ$, so all $G$-orbits have prime power (more precisely, power of $p$) size. Since these all need to sum to $\left|V^\circ\right|=p^n-1$, there is at least one orbit of size $1$. This orbit is a $G$-invariant one-dimensional subspace, and thus is an isomorphic copy of the trivial representation. But this contradicts that $V$ is irreducible.

Answer 2

It is surprising to see that nobody mentioned a beautiful proof of a slightly more general statement presented in Serre's "Linear Representations of Finite Groups". This is Proposition $26$ in $\S8.3$ of Serre's book:

Proposition 26. Let $V$ be a nonzero (not necessarily finite dimensional) vector space over a field $k$ of characteristic $p$ and let $\rho:G\to \mathrm{GL}(V)$ be a representation of a finite $p$-group $G$ on $V$. Then, $V^G\neq 0$, i.e., there exists a nonzero vector $v\in V$ which is fixed by $\rho(g)$ for all $g\in G$.

Proof. Pick any $v\in V\setminus\{0\}$ and let $X$ denote the subgroup (which is just the $\mathbb{F}_p$-vector space) generated by $\{\rho(g)v\mid g\in G\}$. Then for some $n$, $X$ is an $n$-dimensional $\mathbb{F}_p$-vector space on which the finite $p$-group $G$ acts. Since $X$ is the disjoint union of all of its $G$-orbits (which is either a singleton when a representative is $G$-invariant, or a set whose cardinality is a positive power of $p$ otherwise, by the orbit-stabilizer theorem), we deduce that $|X^G| \equiv |X| = p^n \equiv 0 \pmod{p}$. Since $0\in X^G$, $p \leq |X^G| \leq |V^G|$. $\mbox{ }\blacksquare$