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.




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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?




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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. 



Uncertainty principle on finite groups
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.


