# Constituents of induced representation

Let $H < G$ be finite groups with $|G:H|=n$, and let $M$ be an irreducible $FH$-module for some field $F$. Is it always true that all irreducible constituents of the induced $FG$-module $M^G$ have dimension at least $\dim M$? If not, then is it true that the composition length of $M^G$ is at most $n$?

For complex representations, these statements follow from Frobenius Reciprocity. But do they remain true for modular representations? I am particularly interested in the case when $F$ is a finite prime field.

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## 2 Answers

I am assuming you mean "compositon factor" when you speak of irreducible constituent. In the algebraically closed case you seem to be asking the following question in the first part, phrased in terms of Brauer characters: Let $\phi$ be a Brauer character of $H,$ and $\psi$ be a Brauer character of $G.$ Let $\alpha$ be the Brauer character of the projective indecomposable of $H$ corresponding to $\phi$ and $\beta$ be the Brauer character of the projective indecomposable of $G$ corresponding to $\psi.$ Can we have $\langle \beta|_{H}, \phi \rangle \neq 0$ when $\psi(1) < \phi(1)?$ Put this way, I don't immediately see why not, so I think the answer to the first question may be no, but that is not a proof, and I may have missed something.

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@Geoff: I'm also skeptical about transferring this kind of char 0 result to char p>0, I strongly suspect the answer to both questions is no. The first type of likely counterexample which comes to mind involves a rank 2 Chevalley group over a prime field, with a maximal proper parabolic subgroup whose irreducibles include those lifted from a rank 1 Chevalley subgroup and can vary in dimension up to the prime. Inducing one of these from the parabolic should give a filtration with some composition factor of dimension 1, etc. But I'd have to check all the details. – Jim Humphreys Oct 25 '12 at 23:14
@Jim: I agree. In fact I ws in the middle of writing something similar up when I lost the text, and it's getting late here, so I won't start again. I think is would suffice to find $H$ so that $H/Op(H)$ is perfect, and such that the projective cover of the trivial module for $G$ does not restrict to $H$ as just a multiple of the projective cover of the trivial module or $G$. Then you could take $\psi$ trivial and $\phi$ of degree greater than $2$ in what I wrote ($\phi$ and $\psi$ were supposed to be irreducible). – Geoff Robinson Oct 25 '12 at 23:29
I meant greater than or equal to $2$! – Geoff Robinson Oct 25 '12 at 23:29
Thanks Jim, Geoff. Using these ideas I had no difficulty in finding a couterexample to the first question. A maximal parabolic subgroup of $G={\rm SL}(3,3)$ has a 2-dimensional representation that has a 1-dimensional composition factor when induced to $G$. But I have not found any example of this type that comes anywhere near to providing a counterexample to the second question, and it may turn out to be more difficult to refute that, even if there is no reason to believe it to be true. – Derek Holt Oct 26 '12 at 10:05

Write $M^G$ as a direct sum of translated copies of $M$, one for each coset in $G/H$, that are permuted by the action of $G$. This defines a projection $\varphi : M^G \to M$ that looks only at the copy corresponding to the trivial coset. This projection is $H$-equivariant, so the image of an irreducible subrepresentation of $M^G$ is either $0$ or all of $M$.

Let $v$ be a nonzero element of the irreducible subrepresentation. Then it must have a nonzero projection onto the copy of $M$ corresponding to some coset, $gH$. Then $g^{-1} v$ has a nontrivial projection onto the copy of $M$ corresponding to $H$, so the image of $\varphi$ is not $0$, so it is all of $M$, so the dimension is at least $\operatorname{dim} M$.

This method only gets irreducible subrepresentations, not composition factors.

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Do you assume H normal in G in your answer? Since otherwise the typical direct summand of the restriction to H will be $Ind_{H \cap H^g}^H M^g$ and this need not be semi-simple, and the restriction of $M^g$ to $H \cap H^g$ need not be irreducible. – labirintas Oct 25 '12 at 18:03
I believe I've fixed labirintas's objection. – Will Sawin Oct 25 '12 at 22:08
I think this is exactly the argument "by Frobenius reciprocity", which OP mentions in his question. – vytas Oct 27 '12 at 17:29
OK that makes sense. I thought the question might be about subrepresentations rather than composition factors, in which case OP wouldn't know this argument, in which case it would be helpful. – Will Sawin Oct 27 '12 at 18:00