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I am curious about the irreducible representations $\rho: D_{2p} \rightarrow GL_n(\mathbb{Q})$ of dimension at most $p-1$, not the real or complex representations. My mind is occupied with these two questions, especially the first question:

Given a prime number $p$, what are the faithful irreducible representations of the dihedral group $D_{2p}$? What are the irreducible representations of the dihedral group $D_{2p}$?

For $p=5$, do we have irreducible faithful representations of dimensions less than $4$? How can we write down the $4\times4$ rational matrices corresponding to $D_{10}$?

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Over $\mathbb{Q}$, $\mathbb{Z}/p$ has two irreducible representations: the trivial one, and one of dimension $p-1$. Indeed, the group algebra is $\mathbb{Q}[t]/(t^p-1) = \mathbb{Q} \times \mathbb{Q}[t]/(1+t+\ldots+t^{p-1}) = \mathbb{Q} \times \mathbb{Q}(\zeta_p)$ by the irreducibility of the cyclotomic polynomial.

A faithful representation of $D_{2p}$ must have non-trivial restriction to $\mathbb{Z}/p$, so must contain the non-trivial representation of this group, so must have dimension $\geq p-1$.

To write an action by $(p-1 \times p-1)$-matrices concretely, take $V = \mathbb{Q}(\zeta_p)$ with evident basis $1,\zeta_p,\ldots,\zeta_p^{p-2}$, let $\mathbb{Z}/p \subset D_{2p}$ act by multiplication by $\zeta_p$, and let $\mathbb{Z}/2 \subset D_{2p}$ act by complex conjugation.

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  • $\begingroup$ Can we say something special about the dimension of irreducible representations? $\endgroup$ Mar 25, 2021 at 21:20
  • $\begingroup$ Could you please introduce a reference containing these kinds of statements "Over $\mathbb{Q}$, $\mathbb{Z}/p$ has two irreducible representations: the trivial one, and one of dimension $p-1$"? I mean a reference about representations $\rho: G \rightarrow GL_n(\mathbb{Q})$. $\endgroup$ Mar 25, 2021 at 21:33
  • $\begingroup$ @NeoTheComputer, I don't know what general reference @‍sdr might have had in mind, but, in this case, the non-trivial representations of $\mathbb Z/p\mathbb Z$ form a single orbit under $\operatorname{Aut}(\mathbb Q(\zeta_p)/\mathbb Q)$, so the result about two irreducible representations follows. $\endgroup$
    – LSpice
    Mar 25, 2021 at 22:02

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