# Invariant subpaces in the action of the $p-$ group on finite field vector spaces

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.

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.