TITLE: Restriction to a normal subgroup
QUESTION [13 upvotes]: More exam preparation.

Let $A$ be a normal subgroup of a finite group $G$ and $V$ an irreducible representation of $G$. Show that either $\text{Res}_A^G V$ is isotypic (a sum of copies of one irreducible representation of $A$) or that $V$ is induced from some proper subgroup of $G$.

Now, normally when I am asked to prove $P \vee Q$, I see a reasonable path to proving either $\neg P \Rightarrow Q$ or $\neg Q \Rightarrow P$, but I don't see how to do either here. It seems hard to make use of $\neg Q$ as a hypothesis, and I don't see how to relate $\neg P$ as a hypothesis to $Q$ (mainly because I don't see how to produce the needed subgroup $H$). The material most likely to be relevant to this question is, I guess, Mackey restriction or Mackey irreducibility, but I don't quite see how either of these can be applied.

REPLY [5 votes]: Someone should mention Clifford's theorem here. Also, as long as you are working with representations over a field, the result holds in any characteristic. Clifford's theorem tells you that if $U$ is an irreducible submodule of ${\rm Res}^{G}_{A}(V)$, then there is a set $S$ of elements of $G$ such that ${\rm Res}^{G}_{A}(V) = \oplus_{s \in S} Us.$
The proof is to note that $Ug$ is an $A$-submodule of ${\rm Res}^{G}_{A}(V)$ for each $g \in G$, and is clearly irreducible. Now $\sum_{g \in G}Ug$ is a $G$-submodule of $V$, so must be all of $V$, as $V$ is irreducible. Since this is a sum of irreducible $A$-modules, it may be written as a direct sum of some of them. Now $\{ g \in G: Ug \cong U\}$ (as $A$-module) is a subgroup of $G$ containing $A$, which is the inertial subgroup of $U$, and which I will denote by $I$. Clearly $Ug \cong Uh$ as $A$-module if and only if $gh^{-1} \in I$. Hence the number of distinct isomorphism types of irreducible $A$-module which appear in the direct sum is $[G:I].$ Also the inertial subgroup of the irreducible $A$-module $Ux$ is $x^{-1}Ix$ for each $x \in G$.
Let $W$ be an irreducible submodule of ${\rm Res}^{G}_{I}(V)$ such that ${\rm Res}^{I}_{A}(W)$ has a submodule isomorphic to $U$. The argument above, but applied within $I$, shows that ${\rm Res}^{I}_{A}(W) \cong eU$ for some integer $e$, where $eU$ denotes the direct sum of $e$ copies of $U$. Now we can apply Mackey's theorem to ${\rm Res}^{G}_{A}({\rm Ind}_{I}^{G}(W)).$ Since $Ux \not \cong U$ as $A$-module when $x \not \in I$, we see that ${\rm Res}^{G}_{A}({\rm Ind}_{I}^{G}(W)) \cong e(U_1 \oplus U_2 \ldots \oplus U_t)$, where $t = [G:I]$. But ${\rm Res}^{G}_{A}(V)$ must already pick up 
$e(U_1 \oplus U_2 \oplus \ldots \oplus U_t)$, so ${\rm dim}(V) \geq [G:I] {\rm dim}(W).$ However, ${\rm Hom}_{H}(W,{\rm Res}^{G}_{I}(V)) \neq 0$, so that ${\rm Hom}_{G}({\rm Ind}_{I}^{G}(W), V) \neq 0$. Thus the irreducible module $V$ is an epimorphic image of ${\rm Ind}_{I}^{G}(W)$. Since we already have ${\rm dim}(V) \geq {\rm dim}({\rm Ind}_{I}^{G}(W))$, we see that  $V \cong {\rm Ind}_{I}^{G}(W).$