# When an irreducible representation is an induced representation

So I've been trying to answer this exercise to much of my frustration:

Let $$G$$ be a finite group and $$S$$ a normal subgroup. Let $$\rho$$ be an irreducible representation of $$G$$ over $$\mathbb{C}$$. Prove that either the restriction of $$\rho$$ to $$S$$ has all its irreducible components $$S$$-isomorphic to each other, or there exists a proper subgroup $$H$$ of $$G$$ containing $$S$$ and an irreducible representation $$\theta$$ of $$H$$ such that $$\rho \simeq \text{ind}_H^G(\theta)$$.

Here is my progress so far:

Let $$E$$ be a representation space for $$\rho$$ and $$\chi_\rho$$ the character. We also have the restriction $$\text{res}_S^G(\rho)$$ of our representation $$\rho$$ to $$S$$. $$\text{res}_S^G(E)$$ is the representation space for this restriction.

We pick a simple $$S$$-submodule $$F$$ of $$\text{res}_S^G(E)$$ and realize that because $$E$$ is a simple $$G$$-module, $$\begin{equation} \text{res}_S^G(E)=\sum_i \gamma_i F \end{equation}$$ where $$\{\gamma_i\}$$ is a set of left coset representatives for $$G/S$$. Also note that $$S\trianglelefteq G$$ implies that $$\gamma_i F$$ is an irreducible $$S$$-submodule of $$\text{res}_S^G(E)$$. If these submodules are $$S$$-isomorphic to each other, then we have the first case.

Now suppose otherwise. I am guessing that we are able to find a subgroup $$S\subseteq H\subsetneq G$$ and an irreducible character $$\chi_\theta$$ in the character decomposition of $$\chi_{\text{res}_H^G(\rho)}$$ such that $$\text{ind}_H^G(\chi_\theta)$$ is simple. We would then have $$\begin{equation} \langle\text{res}_H^G \chi_\rho,\chi_\theta\rangle=\langle\chi_{\text{res}_H^G(\rho)},\chi_\theta\rangle\geq 1. \end{equation}$$ By Frobenius reciprocity, we would have $$\langle \chi_\rho,\text{ind}_H^G(\theta) \rangle \geq 1$$, which implies $$\langle \chi_\rho,\text{ind}_H^G(\chi_\theta) \rangle = 1$$ since both characters are irreducible. If $$\theta$$ is a representation corresponding to $$\chi_\theta$$, then $$\rho\simeq \theta$$.

Now it remains to find this subgroup $$H$$. I have a hunch that $$H=\{\sigma\in G: \sigma F \simeq F\text{ as an S-representation space}\}$$. It can be easily shown that $$H\subsetneq G$$ is a proper subgroup. However, I am not sure how to find such an irreducible character $$\chi_\theta$$ with our desired properties. Any hints are appreciated!

Your hunch on $$H$$ is correct. For completeness, I redo the problem.
Let $$\rho: G\to \text{GL}(V)$$ be the representation given in question, consider the canonical decomposition $$V = \bigoplus_i V_i$$ when $$V$$ is viewed as an $$S$$-module. $$V_i$$ is a direct sum of isomorphic irreducible representations of $$S$$, and every irr-rep of $$S$$ in this isomorphism class is contained in $$V_i$$.
Because $$S\lhd G$$, $$G$$ permutes $$V_i$$. (Proof: if $$W\subset V_i$$ is a simple $$S$$-module, then normality shows $$gW$$ is also a simple $$S$$-module, and if $$W_1, W_2$$ are $$S$$-isomorphic, then so are $$gW_1, gW_2$$. Summing over all $$W$$ gives the claim.) The action of $$G$$ on $$\{V_i\}$$ is transitive since $$\rho$$ is irreducible.
If there is only one $$V_i$$, then we are done. Otherwise, fix one $$V_1$$, let $$H = \{ g\in G | gV_1 = V_1\}$$ satisfies $$S\subset H$$, $$H\neq G$$. $$\text{Ind}_H^G(V_1)$$ is obviously $$\rho$$. Therefore it remains to prove $$V_1$$ is a simple $$H$$-module. If $$W_0\subset V_1$$ is a proper subspace invariant under $$H$$, then $$W_0$$ contains a simple $$S$$-module $$W$$. Then $$g\in H \iff gV_1 = V_i \iff gW\subset V_1$$ Since $$\rho$$ is irreducible, this forces $$W_0 = V_1$$. Completing the proof.