TITLE: Uncertainty principle on finite groups
QUESTION [12 upvotes]: 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.

REPLY [10 votes]: 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 $G$.

REPLY [7 votes]: 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.