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

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?

• $|G|=\sum_i \dim(V_i)^2$ does not hold in general. As far as I know, it is true for semisimple algebra over an algebraically closed field.
– Bach
Sep 14, 2020 at 20:35
• Here's another proof: math.stackexchange.com/questions/1211369/…
– Ken
Sep 26, 2020 at 6:39

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.

• A very clean method of thinking of it. Nov 19, 2013 at 7:47
• @Alexander Gruber I dont understand one thing. Why do you say that $|V^{\circ}|=p^n-1$? Nov 1, 2014 at 21:48
• @alexander You're assuming $K = \mathbb{F}_p$, but he only said $K$ is of characteristic $p$. Mar 31, 2015 at 2:05
• @Eric, see the comment by Mariano on this. $K$ contains $\mathbb{F}_p$ as the prime subfield, so we can define $$W=\mathrm{Span}_{\mathbb{F}_p}(\mathcal{O}(v))$$ where $\mathcal{O}(v)$ is the orbit of some non-zero $v\in V$. Such $W$, as a subset of $V$, is $G$-invariant, and so it is a representation of $G$ over $\mathbb{F}_p$, but it is not necessarily a subspace of $V$, and therefore might not be a sub-rep. I'm stuck here. Thoughts? Feb 19, 2016 at 23:14
• OK, I think I figured it out Feb 19, 2016 at 23:52

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 $$0\in X^G$$, $$p \leq |X^G| \leq |V^G|$$. $$\mbox{ }\blacksquare$$