Stanford math qual: Linear rep of a $p$-group over $\Bbb{F}_p$ fixes a line pointwise I'm trying to solve a qual question that goes as follows.


Let $H$ be a $p$-group and $V$ a finite dimensional linear representation of $H$ over $\Bbb{F}_p$. Then there is a vector $v \in V$ that is fixed pointwise by $H$.


When $H$ is cyclic of order $p$, this is easy to do. We choose $h\in H$ a generator of order $p$. As an endomorphism of $V$, $h$ satisfies $ (h-1)^p  = h^p - 1 = 0.$ Hence there is $v \in V$ such that $hv = v$. Since $H = \langle h\rangle$, we are done. However, if $H$ is not cyclic, how can I go about doing this? 
My idea when $H$ is not cyclic is to choose an element of order $p$ in the center of $H$, but that idea doesn't seem to go anywhere. Also, what is special about $\Bbb{F}_p$ (and not just any field of characteristic $p$) that I'm not using here? There's a second part  to the question that asks one to show that the same is true over $\overline{\Bbb{F}_p}$. 
 A: The number of elements in $V$ is a power of $p$. Because $H$ is a $p$-group, its orbits on $V$ have sizes that are also powers of $p$. Therefore the number of singleton orbits is a multiple of $p$. Because the zero vector forms a singleton orbit, there must be others.
Part two? There are finitely many entries in the matrices representing elements of $H$. Because all those matrix entries are algebraic over $\Bbb{F}_p$ they all belong to some finite subfield of $\overline{\Bbb{F}_p}$. Proceed as in part one.

An alternative solution (summary of the discussion in the comments with some key contributions from Ben Lim) is to use induction on $|H|$. The base case $|H|=p$ was covered in the question.


*

*We can replace $V$ with any non-trivial subspace stable under $H$, so (as $\dim V<\infty$) w.l.o.g. we can assume that $V$ is an irreducible representation of $H$ (it suffices to find that trivial 1-dimensional representation within an irreducible submodule of the original module).

*Let $h$ be an element of order $p$ in the center $Z(H)$. As deduced in the question $h$ has an eigenvector belonging to eigenvalue $\lambda=1$.

*By Schur's lemma $h$ acts on all of $V$ as a scalar. By the previous bullet that scalar has to be one. 

*If $K$ is the subgroup generated by $h$, then $K\unlhd H$. By the previous bullet we can view $V$ as a linear representation of $H_1=H/K$ by inflation.

*$|H_1|<|H|$, so by induction hypothesis $V$ has a 1-dimensional subspace $U$ such that $H_1$ acts trivially on it.

*Clearly $H$ acts trivially on $U$, and we are done.

A: You can think of the first argument given by Jyrki Lahtonen as an application of the $p$-group fixed point theorem, which just says that if a (finite) $p$-group $G$ acts on a (finite) set $X$ then
$$|X^G| \equiv |X| \bmod p$$
and in particular if $|X|$ isn't divisible by $p$ then the action has a fixed point. It's surprisingly useful to have a name for this result; in the linked post I give a whole bunch of applications of it. 
