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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.

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For reference, this is Isaacs's CToFG, Th 6.11 (p.82) for the proof, and Cor 6.12 for the nice statement. – Jack Schmidt Jun 3 '11 at 22:57
up vote 7 down vote accepted

Looking things up from a textbook feels like cheating, but...

Assuming that the base field is nice (alg. closed, char not a factor of $|G|$), then there is a result in Jacobson's Basic Algebra II (Corollary 4 in section 5.11. - leading up to Brauer's Theorem on induced characters) stating that, if you restrict an irreducible representation $V$ of $G$ to a normal subgroup $H$, and pick any irreducible $H$-component $M$ of $Res^G_H V$, and let $T$ be the inertial group of $M$, then $V=Ind_T^GN$ for some irreducible representation $N$ of $T$.

So it remains to show that the inertial group $T$ is all of $G$, if and only if the representation $Res^G_H V$ is isotypic. Given that all the twists $M^g$, $g\in G\setminus H$ appear in $V$ as $H$-components, it is clear that: isotypic $\Rightarrow$ $T=G$. The other direction follows from the fact that the sum of the irreducible $H$-components of the form $M^g$ of $V$ is stable under the action of $G$, and hence is all of $V$. The assumption $T=G$ means that $M^g$ is isomorphic to $M$ as an $H$-module for all $g\in G$. Therefore $Res^G_H V$ is an isotypic $H$-module.

I don't know, if this answer is useful to you at all, in the sense that the beef was hidden in that Corollary in Jacobson's book. It just seems to fit the bill. You are correct in that this section is all about Mackey's Theorem.

Wish I could say something more helpful about this. It's been about 20 years since I really did any representation theory :-(

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Sorry, what is the inertial group of $M$? – Qiaochu Yuan Jun 3 '11 at 14:51
The inertial group of $M$ consists of those elements $g\in G$ such that $M^g$ is isomorphic to $M$ as an $H$-module. Here we can probably simplify things by using $g\cdot M$ in place of $M$. IOW let $g$ act on $M$ inside $V$. – Jyrki Lahtonen Jun 3 '11 at 14:57

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 $V \cong {\rm Ind}_{I}^{G}(W).$

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