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Let $G$ be a finite group and $H$ a subgroup. Let $\phi$ be an irreducible character of $H$ and $\mathbb 1$ the trivial character of the trivial subgroup $1$. Let $\langle,\rangle_G$ be the usual inner product of characters of $G$.

I am trying to simplify the inner product $\langle \operatorname{Ind}^G_1 \mathbb 1, \operatorname{Ind}^G_H\phi\rangle_G$ where $\operatorname{Ind}$ is the induction of characters.

By Frobenius reciprocity this is equal to $\langle 1, \operatorname{Res^G_1 \operatorname{Ind}^G_H\phi}\rangle_1$.

My question is

Does this simplify to $[G:H]\phi(1)$?

Would I need to use one of Mackey's theorems to proceed further?

Many thanks.

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    $\begingroup$ Yes, it simplifies to that without anything further. There is just a single irreducible representation of the trivial group, so the restriction just gives a suitable number of copies of that. $\endgroup$ – Tobias Kildetoft May 3 '16 at 11:47
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With the help of Tobias Kildetoft's comments I think I can now provide an answer to my question.

Recall that $\operatorname{Ind}^G_1 1$ is in fact the regular character of $G$ and so $\operatorname{Ind}^G_1 1=\sum_\chi \chi(1) \chi $ where the sum is taken over all the irreducible characters of $G$.

Therefore we have $\langle\operatorname{Ind}^G_11, \operatorname{Ind}^G_H \phi\rangle_G= \sum_\chi\langle\chi,\operatorname{Ind}^G_H\phi\rangle_G\chi(1)=(\operatorname{Ind}^G_H\phi)(1)=[G:H]\phi(1)$.

There was no need to use Frobenius reciprocity nor any of Mackey's theorems.

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