Let $p$ be a prime number, and let $G$ be a finite group whose order is a power of $p$. Let $F$ be a field of characteristic $p$, and $V$ a nonzero vector space over $F$ equipped with a linear action of $G$. Does there exist a nonzero subspace $W \subset V$ such that G acts trivially on W?
I am stuck at this question. Now, a vector space on a finite field is again a finite field (I think so). Now, the finite field $F$ is itself a vector space over $F$. I think the group $G$ acts trivially on $F$. Is this right? Thanks beforehand.
Best Answer
We want to show that $G$ has a fixed point on $V$. In order to do so, we let $0 \neq v \in V$ and consider the $\mathbb{F}_p$-span of the orbit of $v$ under $G$, let's call this space $V'$. This is a finite dimensional vector space over $\mathbb{F}_p$, thus $|V'| = p^n$ for some $n$. The group $G$ acts on $V'$ and we have a partition $$V' = \bigsqcup O_i $$ into $G$-orbits. Note that either $|O_i| = 1$, which means that the single element in $O_i$ is a fixed point or $p$ divides $|O_i|$ (orbit-stabilizer-theorem). The identity $$ \sum_{i} |O_i| \equiv 0 \text{ (mod }p)$$ yields that $\{i \:|\: |O_i| = 1\}$ is either empty or contains at least $p$ elements. However, since $0 \in V'$ is a fixed point, it follows that this set is not empty and thus there exists at least one non-zero fixed point in $V'$. Take $W$ to be the subspace generated by this element and you are done.