$A$ is a finite abelian group, $F$ is a finite field and $F[A]$ is the group algebra of $A$ over $F$.
Let $V$ be an irreducible finite-dimensional $F[A]$-module. Let $\rho : F[A] \rightarrow GL(V)$ be the corresponding representation.
I think I have a proof that $\rho(A)$ must be cyclic, and I'd like to know if the proof is valid. My argument is as follows:
Proof:
Since $V$ is irreducible, we must have by Schur's Lemma that $\text{Hom}_{F[A]}(V,V)$ (the ring of $F[A]$-module-homomorphisms from $V$ into $V$) is a division algebra.
Since $A$ is abelian, $F[A]$ is a commutative ring, so $\rho(F[A])\subseteq\text{Hom}_{F[A]}(V,V)$.
This means that $\rho(F[A])$ is a finite commutative division ring; that is to say, it is a finite field! In particular, its multiplicative group of units $\rho(F[A])^\times$ is cyclic. Since $\rho(A)$ is a subgroup of $\rho(F[A])^\times$, it follows that $\rho(A)$ is cyclic. $\blacksquare$
I'd like to know if I'm deluding myself.