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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

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    $\begingroup$ The dimensions of the simple modules divide the order of the group. Since $pq-p<q^2$, they must all have dimension p. I believe you can also get the simple modules by inducing certain $1$-dim characters of $C_q$ and using Mackey's criterion. $\endgroup$ – Benjamin Steinberg Apr 25 '14 at 15:13
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    $\begingroup$ @NickGill: I think you are misreading the condition. $\endgroup$ – David Hill Apr 25 '14 at 15:18
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    $\begingroup$ Ha! Yes, you're right. sorry, i commented in haste. $\endgroup$ – Nick Gill Apr 25 '14 at 15:36
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    $\begingroup$ Your $d=2$ proof is broken: $1^2+7^2=2 \times 5^2$; $7^2+17^2 = 2 \times 13^2$. In general, if $a^2+b^2=p$, then $(a^2-2ab-b^2)^2 + (a^2+2ab-b^2)^2=2 p^2$. (Found by playing with Gaussian integers.) $\endgroup$ – David E Speyer Apr 25 '14 at 17:34
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    $\begingroup$ You write that you "ignored the fact that $\mathbf Z[\sqrt{2}]$ is not a unique factorization domain," but in fact it is a unique factorization domain! $\endgroup$ – KConrad Apr 8 '15 at 1:46
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The $p$ dimensional simples are the inductions of the 1-d reps of $C_q$. That is, the generators of order $p$ and $q$ act by the matrices

$A=\begin{bmatrix} 0 & 0 & \cdots &0 & 1\\ 1 & 0 & \cdots &0& 0\\ 0& 1 & \cdots &0& 0\\ \vdots&\vdots &\ddots&\vdots &\vdots\\ 0 & 0& \cdots &1& 0 \end{bmatrix}\,$ and $\,B=\begin{bmatrix} \zeta & 0 & \cdots &0& 0\\ 0& \zeta^b & \cdots &0& 0\\ \vdots&\vdots &\ddots&\vdots &\vdots\\ 0 & 0& \cdots &\zeta^{b^{p-2}}& 0\\ 0 & 0 & \cdots &0 & \zeta^{b^{p-1}}\\ \end{bmatrix}\,$
for $\zeta$ any $q$th root of unity, and $b$ an element with multiplicative order $p$ in $\mathbb{Z}/q\mathbb{Z}$. There are $d$ distinct ones, since these will be isomorphic if the same root of unity appears in the diagonal of $B$ in both.

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I don't think you mean $d = \frac{q-1}{p}$ in your presentation. I think (as Ben Webster has already implicitly noted) you want $yxy^{-1} = x^{b}$ where $b^{p} \equiv 1 \not \equiv b$ (mod $q$). Ben's answer is correct, of course, but I'll try to outline a solution at the level that students who've just seen Maschke's theorem (and maybe Schur's Lemma) could see without much further theory. So, irreducible complex characters of Abelian groups have degree $1$ by Schur's Lemma. There are $p$ irreducible characters of $G$ of degree $1$ with $\langle x \rangle$ in their kernel. Let's show directly that an irreducible representation of $G$ which does not contain $x$ in its kernel has degree at least $p.$ (Then a dimension count forces all non-linear irreducible characters to have degree exactly $p$). This amounts to explicitly constructing directly the induced modules Ben discusses, but assuming no extra theory a priori. Let $V$ be the associated module. There must some non-zero vector $v \in V$ and some $\omega \neq 1$ such that $vx = \omega v.$ Then $\omega$ must be a primitive $q$-th root of unity as $x^{q} = 1.$ Now $vyxy^{-1} = vx^{b} = \omega^{b}v.$ Hence $vy$ is an eigenvector of $x$ with eigenvalue $\omega^{b}.$ Repeating the argument several times, we see that $v,vy,vy^{2},\ldots vy^{p-1}$ are eigenvectors of $x$ such that $vy^{j}$ is associated to eigenvalue $\omega^{b^{j}}.$ Hence $V$ has dimension at least $p,$ as claimed.

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I like to first think on $\mathbb{Q}G$. For a subgroup H of G let $\widehat{H}=\frac{1}{|H|} \sum_{h\in H} h$. This is an idempotent of $\mathbb{Q}G$ and it is central if and only if $H\unlhd G$. In this case $\mathbb{Q} G \widehat{H}\cong \mathbb{Q} G/H$. For example, $\mathbb{Q} G \widehat{G'}$ is commutative and it is easy to see that every simple component of $\mathbb{Q} G(1-\widehat{G'})$ is non-commutative. If $G=C_p\rtimes C_q$, as in the question, then $G'=C_p$ and hence $\mathbb{Q} G \widehat{C_p} \cong \mathbb{Q} C_q \cong \mathbb{Q} \oplus \mathbb{Q}(\zeta_q)$, where $\zeta_n$ represents a primitive $n$-th root of unity. Moreover, if $e=1-\widehat{C_p}$ then $A=\mathbb{Q} G e$ is a simple algebra of degree $q$. Actually it is isomorphic to a cyclic algebra $(\mathbb{Q}(\zeta_p)/F,\sigma,1)\cong M_q(F)$, where $F$ is the fixed field by the action given on $\zeta$ as the action of $C_q$ on $C_p$. This shows that $\mathbb{Q} G$ is isomorphic to $\mathbb{Q}\oplus \mathbb{Q}(\zeta_q) \oplus M_q(F)$. Using this one can express $\mathbb{C}G$ as a direct product of simple algebras and then the irreducible representations of $G$ follows at once. This method is extended in great generality in [A. Olivieri, Á. del Río, J.J. Simón, On monomial characters and central idempotents of rational group algebras Communications in Algebra, 32 (2004), no. 4, 1531-1550. DOI: 10.1081/AGB-120028797]

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