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I am trying to construct the induced representations of the dihedral group $G=D_p$, $|D_p|=2p$, if I take the subgroup $H=\langle r \rangle \cong C_p$, which is generated by the rotations. I have problems as I try to understand the general way of the construction, so I try it with this example. It's not a homework!

If $W$ is a $\mathbb{Q}H$-module, then the induced representation is defined as $\mathrm{ind}^G_H (W):=W \otimes_{\mathbb{Q}H} \mathbb{Q}G$. In this example $W$ has to be $\mathbb{Q}$ or $\mathbb{Q}(\zeta_p)$, because these are the simple $\mathbb{Q}H$-modules. For $k=\dim(W)$ and $l=[G:H]$ I can choose a $\mathbb{Q}$-basis $w_1,\cdots , w_k$ of $W$ and a $\mathbb{Q}H$-basis $g_1,\cdots , g_l $ of $\mathbb{Q}G$. Then $$w_1 \otimes g_1,\cdots ,w_k \otimes g_1,\cdots w_k \otimes ,g_l $$ has to be a $\mathbb{Q}$-base of $\mathrm{ind}^G_H(W)$ and $\dim(\mathrm{ind}^G_H(W))=kl$.

So if I first take $W=\mathbb{Q}(\zeta_p)$, what would the induced representation be? I think, that the dihedral group should have two $1$-dim and one $(p-1)$-dim representations. How can I get the $(p-1)$-dim one?

Somehow I cannot write any comments:

Thanks for your help, but I have to work with representations over $\mathbb{Q}$ and later over $\mathbb{R}$. To get representations over $\mathbb{C}$, I could avoid to use the induced representation and for many groups it was not really difficult to get the irreducible representations without it.

Of course I know Maschke's theorem, but it gives me only the existence of irreducible representations, if e.g. $char(G)=0$, but not how to construct them. I hope anyone has an idea, how to work with $ind(W)$ in this expamle?

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I'm curious: what introductory text works over $\mathbf Q$ and not $\mathbf C$? –  Dylan Moreland Apr 18 '12 at 20:16
    
I don't have an "introductory text", just a general definition of the induced representation over a field $\mathbb{K}$. The representations over $\mathbb{C}$ you can usually get without it, so I try to understand it over $\mathbb{Q}$, or are there many differences? –  bounty Apr 18 '12 at 20:26
    
@bounty: This theory is best over algebraically closed fields. So while you don't need $\mathbb{C}$, you do need to at least work in $\overline{\mathbb{Q}}$ in order to have basic results like Maschke's Theorem (and in fact, your representations will live in this field). –  Brett Frankel Apr 18 '12 at 21:48
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@BrettFrankel, Maschke's Theorem holds for a field of characteristic zero without any restrictions... –  Bruno Joyal Apr 18 '12 at 23:37
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@bounty Of course it does, silly me. I was thinking Schur's Lemma. –  Brett Frankel Apr 18 '12 at 23:39

1 Answer 1

The irreducible $(p-1)$-dimensional $\mathbb{Q}G$-module is not induced from a rational representation of $C_p$. But twice that is. One way of seeing this is to work over $\mathbb{C}$. Over $\mathbb{C}$, the $(p-1)$-dimensional guy decomposes as a sum of $\frac{p-1}{2}$ irreducible 2-dimensional representations. Each is induced from a one-dimensional representation of $C_p=H$. However, if $\chi$ is a non-trivial one-dimensional character of $C_p$, then $\mathrm{Ind}_{G/H}\chi = \mathrm{Ind}_{G/H}\bar{\chi}$. That follows from Clifford theory: $H$ is normal in $G$ and $\chi$ and $\bar{\chi}$ lie in one orbit under the action of $G$ on the irreducible characters of $H$.

Thus, if you induce the sum of all non-trivial one-dimensional representations of $H$ to $G$, you hit each irreducible 2-dimensional complex representation of $G$ twice.

Another way of obtaining the $(p-1)$-dimensional guy through induction is by inducing the trivial representation of $C_2\leq G$. This induction will be $p$-dimensional and will have the trivial $G$-representation as a direct summand (this uses Maschke's theorem, which of course works just fine over $\mathbb{Q}$). The remaining $(p-1)$-dimensional chunk is irreducible over $\mathbb{Q}$, and is the representation you are looking for.

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