I've been teaching some elementary representation theory to undergraduates, and want to provide applications of Maschke's theorem to complex group algebras to present in class. In particular, I'd like to work out the following:

Problem: Let $G$ be a nonabelian group of order $pq$, where $p|(q-1)$. Decompose the group algebra $\mathbb{C}G$.

Now, this group is easy to describe, it is the semidirect product of cyclic groups $$G\cong C_q\rtimes C_p=\langle x,y\mid x^q=y^p,yx=x^dy\rangle,$$ where $d=\frac{q-1}{p}$. A straightforward computation shows that there are $1+d+(p-1)$ conjugacy classes in $G$ (with sizes 1, $p$, and $q$, respectively). Since $C_p\cong G/C_q$, there are $p$ nonisomorphic 1-dimensional simple modules. Therefore, $$\mathbb{C}G\cong \mathbb{C}^{\oplus p}\oplus M_{n_1}(\mathbb{C})\oplus\cdots\oplus M_{n_d}(\mathbb{C}).$$ In particular, we have that $n_1^2+\cdots+n_d^2=pq-p=p^2d$.

I'd like to conclude that $n_1=\cdots=n_d=p$, but it is not clear to me how to prove it.

It is obvious if $d=2$. Indeed, assume $n^2+m^2=2p^2$. Then, without loss of generality $m=p-k\leq p$. Hence, $$n^2=p^2+2pk-k^2.$$ The only solution is $k=0$.

Question: How do you prove this when $d>2$?

I'd prefer to have an elementary number theoretic proof, but if someone provides a construction of the $d$ nonisomorphic simple modules of dimension $p$, that would work too. Of course, if I'm wrong about the decomposition, please let me know.

Final note: If the powers that be decide to migrate this to stackexchange, that's fine. Though, based on the questions I've seen there I think this is a more reasonable place to get an answer.

Edit: In light of David Speyer's comment, there is no number theoretic proof of this decomposition, since none would work for a group of order 55. In my "proof" above, I carelessly ignored the fact that $\mathbb{Z}[\sqrt{2}]$ is not a unique factization domain.

The presentation of the group is slightly off, as pointed out by Geoff Robinson. Anyway, the point was to prove the decomposition without any additional theory. Geoff explains how to do this. Though, I think the simplicity of the representations in Ben's answer can be easily obtained as follows: The eigenvalues for the action of $x$ are distinct, so it is easy to write down projection operators onto the eigenvectors. These, in turn generate the whole module, so it must be simple

isa unique factorization domain! $\endgroup$ – KConrad Apr 8 '15 at 1:46