The problem is this:

Let $G$ be a finite group, $Q$ a normal subgroup of prime index.

(i) Let $\psi = \chi_V$ be the character of the $Q$ representation $V$. Show that either $\psi$ can be extended to $G$, or the induced representation is irreducible.

(ii) Show that if $Q$ is abelian, then every irreducible representation of $G$ is either one-dimensional or $p$-dimensional

For the first part, I've done one direction: If $V = \text{Res}(U)$, then $\text{Ind}(V)$ cannot be irreducible in this case - using Frobenius reciprocity and the Frobenius character formula. However, I'm stuck on the reverse direction.

For the second part, I have a strategy I'm a little bit skeptical of. Basically, I'm thinking you take an irrep $V$ of $G$, and restrict to $Q$. If this is an irreducible $Q$ rep, then we're done. Else, it has an irreducible $Q$ subrep $U$. I'm thinking that if I can show that $q \in Q$, $q \not = e$, then $gU \neq U$. In this case, I'm thinking I can then exhibit ${\rm Ind}(U)$ as a $G$ subrep of $V$, and then I would be done since it must be the whole of $V$. But, I'm pretty skeptical this totally works since I neither used the previous part, not the hypothesis that $Q$ is normal of prime index.

If someone could give me some insight into this question, I would appreciate it. I would also appreciate any general advice on dealing with the relationship between representations of some group $G$, and the representations of subgroups and quotients of $G$.

  • $\begingroup$ Are you familiar with Clifford theory? $\endgroup$
    – Nate
    Mar 25, 2020 at 17:24
  • $\begingroup$ Sadly no, is it relevant? $\endgroup$
    – msm
    Mar 25, 2020 at 19:55
  • $\begingroup$ Clifford theory pertains to induction and restriction to and from normal subgroups, and it's very useful for problems like these. (Try google-ing it) $\endgroup$
    – Nate
    Mar 25, 2020 at 21:45

1 Answer 1


Here are some hints to get you started. (I'm assuming your representations are valued over an algebraically closed field: if not, these statements are false.)

For part (i):

Write $G/Q = \langle \sigma Q\rangle$ for some element $\sigma\in G$. Let $\rho = \mathrm{Ind}_Q^G(\psi)$. By Frobenius reciprocity, $$(\rho, \rho)_G = (\psi, \mathrm{Res}^G_Q(\mathrm{Ind}_Q^G(\psi))_Q = (\psi, \bigoplus_{i = 0}^{p-1}\psi^{\sigma^i})_Q.$$ Here, the final equality follows from Mackey theory. If $\psi\simeq \psi^{\sigma^i}$ for some $i\ne 0$, then, since $\sigma$ has prime order, $\psi\simeq \psi^{\sigma^i}$ for all $i$. Hence, $(\rho, \rho)_G$ is either $1$ or $p$. Now use Schur's Lemma to match each of these cases to one of the two options.

For part (ii):

Your strategy almost works. The additional fact you need is that $U$, as an irreducible finite dimensional representation of an abelian group, is one-dimensional. Now the result follows from part (i), Frobenius reciprocity and Schur's Lemma.


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