Dimension of irreducible module divides the dimension of the algebra?

Question

Fact: $\chi(1)$ divides order of $|G|$ where $\chi$ is an irreducible character of $G$.

Above fact is equivalent to say that if $V$ is an irreducible $A=\mathbb C [G]$ module then $\dim(V)$ divides $\dim(A)$.

I wonder that whether this fact can be proven by modules.(without using character theory)

I also wonder that whether it is true for general case;

If $A$ is an algebra over $\mathbb C$ and $V$ is an irreducible $A$ module then $\dim(V)\mid\dim(A)$ ?

Answer 1

I would doubt there is a general fact. First of all because of Jim's counterexample. Secondly, because it fails in positive characteristic (take $\operatorname{SL}(2,p)$ for $p$ odd, see, e.g. [Curtis-Reiner, (17.17)]).

It is probably worth pointing out that in the special case that $A$ is a semisimple Hopf algebra over a field of characteristic zero, this is called Kaplansky's sixth conjecture. There is a mathoverflow question on that, pointing you to several nice survey articles. I haven't looked at the papers containing partial results. It might be that some of them contain more "character-free" approaches.

Answer 2

If $A$ and $B$ are both algebras and $V$ is an irreducible $A$-module, then $V$ is also an irreducible $A \oplus B$ module (where $B$ acts trivially). As the dimension of $A \oplus B$ is $\dim A + \dim B$ and $B$ is essentially arbitrary the answer is no, $\dim(V)$ needn't divide the dimension of the algebra.