Proving that if $|G| > 1$, then every irreducible $FG$-module has dimension $<|G|$ I think that the thing stopping me from solving this is that I don't fully understand the relationship between $FG$-modules and $F$-vector spaces. I believe $FG$-modules are in correspondence with representations $\psi: G \rightarrow GL_n(F)$, but to be honest I don't know how to interpret $GL_n(F)$. It's the set of $n$ by $n$ matrices over $F$, but I don't see where $n$ comes in.
Can someone give me some direction?
 A: For a field $F$ and a finite group $G$, $R=F[G]$ is a quasi-Frobenius ring, and in particular, $R_R$ contains a copy of every simple right $R$ module.
(Actually, the more specific relevant condition is that it is a left and right Kasch ring.)
Since $F[G]$ is $|G|$ dimensional, the dimension of any such copy would have to be less or equal.
Now since $F[G]$ is Artinian, it has finite composition length. A composition series can start at any one of these simple submodules. Now, the only way for $\dim(S)=|G|$ for one of these simple submodules is if the compostion length is $1$: but that would mean $F[G]$ is a simple $F[G]$ module, implying that it is a field. But when $|G|>1$, the augmentation ideal of $F[G]$ is always nonzero, so it is only a field when $|G|=1$. 
Thus if $|G|>1$, $F[G]$ has composition length $>1$, and therefore $S\neq F[G]$ and has strictly smaller dimension.
A: Try something like this. Let $M$ be an irreducible $FG$ module, that means $M$ an $F$ vector space equipped with a left linear action of $G$ such that it has no non trivial submodule. Let $m \in M$ be a non zero vector. Because of $M$ irreducible , the span of $gm$ for all $g \in G$ is the entire $M$... 
