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…]
Best Answer
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.