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Let $G$ be a group and let $H$ be a subgroup of finite index.

Let $V$ be an irreducible complex representation of $G$ (no topology or anything: $V$ is just a non-zero complex vector space with a linear action of $G$ and no non-trivial invariant subs).

Now consider $V$ as a representation of $H$. Is $V$ a finite direct sum of irreducible $H$-reps?

I am almost embarrassed to ask this question here. It looked to me initially like the answer should be "yes and this question is trivial". If $G$ is finite it is trivial and Clifford theory tells you basically what can happen. Here is another case I can do: if $H$ has index two in $G$ then $V$ is indeed a finite direct sum of irreducibles. For either $V$ is irreducible as an $H$-rep, in which case we're done, or $V$ is reducible, so there's $0\not=W\not=V$ an $H$-stable sub. Say $g\in G$ with $g\not\in H$. One checks easily that $gW$ is $H$-stable, that $W\cap gW$ is $G$-stable, so must be zero, and that $W+gW$ is $G$-stable, so must be $V$. Hence $V$ is the direct sum of $W$ and $gW$. This implies that $W$ is irreducible as an $H$-rep---for if $X$ were a non-trivial sub then the same argument shows $V=X\oplus gX$ but this is strictly smaller than $W\oplus gW=V$.

I thought that this argument should trivially generalise to, say, the case where $H$ is a normal subgroup of prime index. But I can't even do the case where $H$ is normal and $G/H$ has order $3$, because I can't rule out $V$ being the sum of any two of $W$, $gW$ and $g^2W$, and the intersection of any two being trivial.

Either I am missing something silly (most likely!) or there's some daft counterexample. I almost feel that I would be able to prove something if I knew Schur's lemma [edit: by which I mean that if I knew $End_G(V)=\mathbf{C}$ then I might know how to proceed], but in this generality I don't see any reason why it should be true. Perhaps if I knew a concrete example of an irreducible complex representation of a group for which Schur's lemma failed then I might be able to get back on track. [edit: in a deleted response, Qiaochu pointed out that $G=\mathbf{C}(t)^\times$ acting on $\mathbf{C}(t)$ provided a simple example] [final remark that in the context in which this question arose, $G$ was a locally profinite group and $V$ was smooth and I could use Schur's Lemma, but by then I was interested in the general case...]

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What about the way it is done in Labesse-Langlands for the local groups? It seems pretty general, but I don't know how general you need it and what you have in mind –  MBN Mar 16 '11 at 20:54
Heh :-) I looked in there. As far as I can see, they only deal with the case where $H$ has index 2. For they are interested in the case where $F$ is a local field, $G=GL(2,F)$ and $H$ is the subgroup generated by $SL(2,F)$ and the centre; this is normal with quotient a finite group of the form $(Z/2Z)^n$ so they only need the index 2 case and then they can proceed by induction. The proof they give in the index 2 case is similar and probably the same as the one I present above. –  Kevin Buzzard Mar 16 '11 at 21:39
I was lazy to take a look, sorry. I still cannot rid myself of the feeling that I have seen it treated for a general index somewhere, although they were reproducing the L.-L. result and didn't need it in that generality. I am probably wrong and just having a bad recollection. –  MBN Mar 16 '11 at 21:56
This is frustrating. I feel like if a counterexample exists it ought to be relatively easy to construct but I have already spent far too much time trying to do so... –  Qiaochu Yuan Mar 16 '11 at 22:48
@MBN: if one is dealing with smooth irreducible representations of a locally profinite group then one often has Schur's Lemma at one's disposal and one can use Frob Rec to get some sort of hold on End_H(W) for W some non-trivial H-stable subspace of V (if V isn't irred as H-mod) and then go on from there (it's a finite-dimensional C-algebra and one knows a lot about these). It wouldn't surprise me if this could be turned into a proof when Schur is OK. @Qiaochu: I sympathise! I spent a fair amount of time on this this morning. –  Kevin Buzzard Mar 17 '11 at 0:24
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2 Answers

up vote 9 down vote accepted

Since $V$ is irreducible, it is a finitely generated $\mathbb C[G]$-module, any non-zero element is a generator. Since $H$ is of finite index, $\mathbb C[G]$ is a finitely generated $\mathbb C[H]$-module. Hence $V$ is a finitely generated $\mathbb C[H]$-module. Zorn's Lemma implies the existence of an irreducible quotient $W$.

Suppose that $H$ is normal in $G$. (It is probably enough to suppose that $H$ contains a subgroup of finite index, which is normal in $G$.) Let $K$ be the kernel of $V\rightarrow W$, for every $g\in G$ the quotient $V/g K$ is an irreducible $H$-rep, isomorphic to $W^g$. The kernel of the natural map $$V\rightarrow \bigoplus_{g\in G/H} V/g K$$ is $G$ invariant, and hence $0$. So we may inject $V$ into a finite direct sum of irreducible $H$-reps. Choose a smallest subset $X\subset G/H$, such that $\varphi: V\rightarrow \bigoplus_{g\in X} V/g K$ is injective, then $\varphi$ is also surjective.

Edit. Kevin pointed it out that $H$ always contains a subgroup of finite index, which is normal in $G$, and F.Ladisch finished off the general case, i.e without assuming that $H$ is normal in the comments below.

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@Kevin. The comment is idiotic. I was thinking restrict to a normal subgroup, take an irreducible summand and induce it up to $H$. This should be semisimple. –  vytas Mar 17 '11 at 18:26
@Kevin. Sorry, I mean I am an idiot for not noticing that the situation in the bracketed comment always holds. –  vytas Mar 17 '11 at 18:28
Hi Vytas. In fact I was inspired by Guntram's answer to come up with some related argument on the tube home today---I figured that Guntram's "choose $M$ irreducible" could be replaced by "choose $M$ as small as possible to make all relevant intersections $M\cap\Sigma_i g_iM$ equal to either zero or $M$" and then proceed as Guntram does. I didn't check all the details though and then it was my stop :-/ I see how your argument differs from Guntram's though: it's not clear that $V$ has an irred sub but it is clear it has an irreducible quotient. Can you explain "then $\phi$ is also surjective"?? –  Kevin Buzzard Mar 17 '11 at 18:28
Nice and simple argument! Instead of the last lines, you could simply say that $V$ as a submodule of a semisimple $\mathbb{C}[H]$-module of finite length must be itself such a module. –  Frieder Ladisch Mar 17 '11 at 18:42
The normal case is enough: Let $W\leq V$ be a $H$-subrep and $N\subseteq H$ normal in $G$ with finite index. Then $W$ is a direct summand of $V$ as $N$-module, so there is an $N$-linear projection $\pi\colon V\to W$. Averaging over a set of coset representatives of $N$ in $H$ yields an $H$-linear projection (as in the classical proof of Maschke's theorem, this is possible because $|H:N|$ is finite). This shows that every $H$-subrep is a direct summand as desired. –  Frieder Ladisch Mar 17 '11 at 21:28
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Suppose that $H$ is normal in $G$. Write $G=\cup_{i=1}^n g_iH$.

Let $M < V$ be an irreducible $H$-module and write $M_i:=g_iM$. Since $H$ is normal, $M_i$ is an $H$-module, which is irreducible as $M$ is irreducible. Note that $M_i \cap \sum_{j \neq i} M_j$ is either trivial or equal to $M_i$ since it is a submodule of $M_i$.

Now $V=\sum_{i=1}^n M_i$ since the RHS is a $G$-submodule of $V$. By leaving out superfluous $M_i$'s we get that $V$ is a direct sum of certain of the $M_i$'s.

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Maybe this is a dumb question, but is it obvious that $M$ exists? If $V$ is infinite-dimensional, how do we rule out the possibility that the poset of $H$-subspaces of $V$ has no minimal elements? –  Qiaochu Yuan Mar 17 '11 at 13:16
@Qiaochu, Guntram: existence of M is precisely what held me up with this approach. You have to rule out the possibility that V is irred as G-module but is completely ghastly, infinite length, no irreducible subs at all, as H-module. –  Kevin Buzzard Mar 17 '11 at 17:09
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