0
$\begingroup$

Let $G$ be a group and $H$ a subgroup. Let $\psi$ be a character on $G$ and $\phi$ be a character on $H$. The Frobenius reciprocity tells us that ${\displaystyle \langle \operatorname {Ind} _{H}^{G}\psi ,\phi \rangle _{G}=\langle \psi ,\operatorname {Res} _{H}^{G}\phi \rangle _{H}}$. Now suppose $\psi$ is an irreducible character on $H$ and $\phi$ is an irreducible character on $G$. Does $\operatorname {Res} _{H}^{G}\phi$ need to be exactly one of the irreducible representations of $H$ or can it be a multiple of it? Or can it even be the sum of different irreducible characters of $H$? The latter seem impossible to me but I don't understand why.

$\endgroup$
1
  • $\begingroup$ The coefficients can take all values and there can be any number of non-isomorphic constituents. I am only giving this as a comment because I am not providing examples. $\endgroup$ Jul 29, 2020 at 10:03

1 Answer 1

1
$\begingroup$

Firstly, I think you should have $\psi$ is a character of $H$ and $\phi$ a character of $G$ in order for your inductions and restrictions to make sense.

Assume that $G$ is finite and $H$ is the identity subgroup. Then $H$ only has one irreducible representation: the trivial one. Now let $G$ be a nonabelian group and take an irreducible representation of dimension greater than $1$ (eg the standard one in $S_3$). Then the restriction has no choice but to be $\dim \phi$ copies of the trivial one when restricted to $H$.

If we stick with $G=S_3$ and the standard $2$-dim representation, then its restriction to any subgroup of order two is the regular representation of $C_2$, i.e. the trivial one plus the nontrivial one so different constituents can also occur.

$\endgroup$
2
  • $\begingroup$ Thank you. What is the standard 2-dim representation of $S_3$? $\endgroup$
    – roi_saumon
    Jul 29, 2020 at 12:32
  • $\begingroup$ Simply put, its the only 2-dimensional irreducible representation of $S_3$. See groupprops.subwiki.org/wiki/… for the actual representation $\endgroup$
    – Matt B
    Jul 29, 2020 at 15:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .