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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 $\mathbf C$?2012-04-18
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    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?2012-04-18
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    @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).2012-04-18
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    @BrettFrankel, Maschke's Theorem holds for a field of characteristic zero without any restrictions...2012-04-18
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    @bounty Of course it does, silly me. I was thinking Schur's Lemma.2012-04-18
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    @Brett: It appears you pinged the wrong person. OP is pinged automatically of every comment to his/her question BTW. :)2012-04-19
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    Silly mistake #2 for the day. No point in re-pinging bruno now though, since he already knows the answer.2012-04-19

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