Let $G$ be a finite group, $H$ a normal subgroup of $G$ such that $G/H$ is cyclic. Let $V$ be an irreducible representation over an algebraically closed field $k$, with $char(k)=0$.

Is it true that the restriction of $V$ to $H$ is a direct sum of irreducible pairwise non-isomorphic representations of $H$?

What if $G$ is profinite, $H$ is open in $G$ (with cyclic quotient and $V$ is an $\ell$-adic continuous $\overline{\mathbb{Q}_\ell}$-representation?


1 Answer 1


Yes it is true that the restriction of $V$ to $H$ is a direct sum of pairwise non-isomorphic representations of $H$ in the finite group case. I have no idea what happens in the profinite case.

I am only going to give an outline argument. By Clifford's Theorem, $V$ is the sum of irreducible $H$-representations. If two of these are isomorphic, then by replacing $G$ by its subgroup that stabilizes that representation, we may assume that all of the summands are isomorphic to a single $H$-representation $W$.

Let $g \in G$ be such that its image generates the cyclic group $G/H$. Then the automorphism of $G$ induced by $g$ stabilizes $W$. Let $W(H)$ be the image of $W$ in ${\rm GL}_n(k)$ for some $n$. Then we can find a matrix $\alpha \in {\rm GL}_n(k)$ that effects the equivalence of $W$ with $W^g$, so $\alpha$ normalizes $W(H)$ and induces the same automorphism as $g$. Now (this is where we use the algebraic closure of $k$ - the result is not true without that assumption) we can multiply $\alpha$ by a scalar matrix to make $\langle W(H),\alpha \rangle$ the image of an extension of $W$ to $G$.

In fact, if $|G/H|=m$, then we can multiply $\alpha$ by any scalar matrix of order dividing $m$, and still have an extension of $W$ to $G$, and it is not hard to see that these $m$ extensions are mutually inequivalent.

So we have shown that, in this situation, $W$ extends to $m$ distinct irreducible representation of $G$. By Frobenius Reciprocity, the induced representation $W^G$ has each of these representations as a component, so it is the sum of these $m$ representations. But our original represenation $V$ of $G$ would also need to be a component of $W^G$, which is not possible, so we have a contradiction.

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    $\begingroup$ This result for finite groups is proved in Theorem 2.14 Chapter III of Feit's book "The Representation Theory of Finite Groups". $\endgroup$
    – Derek Holt
    Commented Jun 1, 2017 at 10:00

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