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For a finite group $G$ with normal subgroup $H$, the induced representation $\text{Ind}_H^G(1)$ decomposes as a sum of irreducibles with the multiplicities equal to the dimensions, because it is is the pullback of the right regular representation of $G/H$. For a subgroup which is not normal, this need not be true. For example, take $G=S_3$ and $H$ a subgroup of order $2$. The index $3$ can only be written as a sum of squares as $1^2+1^2+1^2$, but $S_3$ only has two distinct $1$ dimensional representations.

Is the converse true? That is, if $\text{Ind}_H^G(1)$ decomposes into irreducible representations with multiplicity equal to the dimension, is $H$ necessarily normal in $G$? If not, is there some other characterization of subgroups which have this property?

The question arises in generalizing the 'Uncertainty Principle' for finite abelian groups $$ |G|\le \sharp \text{supp}(f)\cdot \sharp\text{supp}(\hat f), $$ to finite groups in general, and when the corresponding inequality is tight.

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    $\begingroup$ You may know this, but Theorem 2 in arxiv.org/pdf/math/0608702v2.pdf shows that the kind of uncertainty principle you quoted holds for compact (in particular finite) groups, appropriately reworded. $\endgroup$
    – Gian Maria Dall'Ara
    Commented Jun 14, 2013 at 8:00

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Yes, it is. If for any irreducible $\rho$, $\langle\text{Ind}_H^G(1),\rho\rangle$ is either 0 or dim $\rho$, then $H$ is the intersections of $\ker \rho$ for those $\rho$ for which this inner product is not 0, and intersections of kernels of irreducible characters are precisely the normal subgroups of $G$.

Indeed, by Frobenius recoprocity, $$ \langle \text{Ind}_H^G(1),\rho\rangle = \langle 1,\text{Res}_H\rho\rangle =\dim\rho\Longrightarrow H\leq \ker\rho. $$ This gives you the inclusion $H\leq \bigcap \ker\rho,$ the intersection running over those $\rho$ with non-trivial inner product with $\text{Ind}_H^G1$. To get equality, notice that the hypothesis on the inner products implies that the dimensions of $\text{Ind}_H^G1$ and of $\text{Ind}_K^G 1$, where $K=\bigcap \ker\rho$, are the same.

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  • $\begingroup$ I see that containment in the kernels is clear, how do you get equality? $\endgroup$
    – Stopple
    Commented Jun 13, 2013 at 17:29
  • $\begingroup$ By comparing dimensions. The hypothesis implies that $\dim(\text{Ind}_H^G 1) = \dim(\text{Ind}_K^G 1)$, where $K=\bigcap \ker\rho$, the intersection taken over those $\rho$ with non-zero inner product with $\text{Ind}_H^G1$. $\endgroup$
    – Alex B.
    Commented Jun 13, 2013 at 19:33
  • $\begingroup$ I have edited the answer. You can also observe directly that the hypothesis gives an isomorphism of representations $\text{Ind}_H^G1$ and $\text{Ind}_K^G1$, where $K$ is the intersection of kernels. $\endgroup$
    – Alex B.
    Commented Jun 13, 2013 at 19:42
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Yes. If an irreducible representation $U$ occurs in the induced representation with multiplicity $\dim(U)$, then it follows by Frobenius Reciprocity that the trivial representation occurs with multiplicity $\dim(U)$ in the restriction of $U$ to $H$, and so $H$ acts trivially on $U$. So your condition implies that $H$ acts trivially on the induced representation, which is only the case if $H$ is normal in $G$.

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