Let $P$ be a finite $p$-group, i.e. $|P|=p^n$ for some prime $p$. Let $F$ be a finite field of $p$ elements. Show that every simple $FP$-module trivial, where $FP$ is a group ring (which is actually a division ring because $F$ is a field).
Definition: A trivial $FP$-module is a one dimensional $F$-vector space where all elements of $G$ acts as identity.
My attempt:
Let $M$ be a simple $FP$-module.
I know $M$ is simple if and only if it can be generated by every non-zero element, i.e. $\forall 0\neq\alpha\in M$, $M=(FP)\alpha$.
If $M$ is a one dimensional $F$-vector space, then we are done. So, we may assume $M$ is not a one dimensional $F$-vector space, which implies there exists some $e_G\neq g\in G$ and some $m\in M$ such that $g\cdot m\neq m$. We want to construct a non-zero proper submodule of $M$, which will contradict with the simplicity of $M$ and then we are done. To this end, I consider the cyclic module $(FP)(g\cdot m)$ as the condition suggests. And I found that $FP=FP(g^{-1})$, which leads to $(FP)(g\cdot m)=[FP(g^{-1})](g\cdot m)=FP[(g^{-1})(g\cdot m)]=FP(e_G\cdot m)=FP(m)$. But then I just get stuck here. I didn't (and didn't know how to) use the conditions that $|P|=p^n$, $|F|=p$ and $g\cdot m\neq m$.
My question is how to use these conditions to show that $FP(m)$ is a non-zero proper submodule of $M$? Or alternatively, can we use other ways to find a a non-zero proper submodule of $M$?
Thanks for help.
Best Answer
A general fact that is useful to know. Say $|P| = p^n$ with $p$ prime, and let $F$ be a finite field with $p$ elements.
To prove this lemma, look at the orbits of $P$ on $V$.
In the case where $V$ is irreducible, the result you want to prove follows from the lemma.