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!


1 Answer 1


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.


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