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Let $G$ be a finite group, and $H$ be a subgroup of $G$.

If I am not mistaken, we have

$$Ind_H^G \,\,1_H=1_G+\sum_{\chi\neq 1} n_\chi \chi$$

where $\chi$ are irreducible representations.

My question is, what representations are these?

I guess is should be a proper subset (if $H$ is a proper subgroup) of the irreducible representations of $G$, but I'm not too sure of that either.

And if that's correct, how do I know which $\chi$ occur in the decomposition above.

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2 Answers 2

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One way to answer this is given by what is called Frobenius reciprocity (for simplicity, I am assuming that we are working over the complex numbers here).

This states that the constituents of the induced representation are precisely those which, when restricted to the subgroup, have the trivial representation as a constituent.

The above also works for any other irreducible representation of the subgroup, and even says what the multiplicities are (being the same as the multiplicity when one restricts).

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  • $\begingroup$ Can you give me a reference for your answer? Preferably some textbook. $\endgroup$ Nov 20, 2019 at 6:03
  • $\begingroup$ @Krish Frobenius reciprocity is covered in basically any treatment of representation theory. $\endgroup$ Nov 20, 2019 at 16:17
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There is a nice interpretation of $1_{H}^G$ in terms of permutation characters. Let $X$ be a set on which $G$ acts. Then the character $\chi$ of the permutation action is given by the following formula: $\chi(g)=\#\{$fixed points of $g$ on $X$$\}$. Now, if $G$ acts transitively on $X$ and $H=H_x$ is the stabilizer subgroup of $G$ on one of the points $x \in X$, then $\chi=1_{H}^G$.

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    $\begingroup$ Maybe it is worth pointing out that a nice way to get such a set is by taking the cosets of $H$ in $G$. $\endgroup$ Sep 23, 2016 at 12:09
  • $\begingroup$ Yes, absolutely. In addition, the kernel of that action is $ker(1_H^ G)=core_G(H)$, the insection of all $G$-conjugates of $H$. Thanks. $\endgroup$ Sep 23, 2016 at 12:12

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