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I know that Frobenius reciprocity helps us to solve this problem, but I don't know why: Let $ G $ be a transitive finite permutation group with permutation character $ \pi $. If $\chi $ is an irreducible $ \mathbb{C} $-character, then it's degree is at least $ \langle\pi, \chi\rangle_G $. when does equality hold?

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  • $\begingroup$ If $H$ is the stabilizer in $G$ of a point, then $\pi$ is the induced character $1_H^G$, where $1_H$ is the trivial character of $H$. $\endgroup$ – Derek Holt May 24 '15 at 11:09
  • $\begingroup$ You should either accept Nicky Hekster's answer or, if there is something that you don't understand, then say what it is. Similarly for math.stackexchange.com/questions/1295076 $\endgroup$ – Derek Holt May 24 '15 at 17:42
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Derek is right, let $G$ act transitively on a set $\Omega$ and put $H=G_\alpha$, for some $\alpha \in \Omega$, then the permutation character $\pi=1_H^G$, the trivial character of $H$ induced to $G$. This is a standard textbook theorem. And indeed, by Frobenius Reciprocity, if $\chi \in Irr(G)$, then $[\chi, \pi]=[\chi, 1_H^G]=[\chi_H,1_H] \leq \chi(1)$.

Equality occurs if and only if $H \subseteq ker(\chi)=\{g \in G : \chi(1)=\chi(g)\}$: if $H \subseteq ker(\chi)$, then $\chi(h)=\chi(1)$, so $\chi_H=\chi(1)1_H$. Conversely, if $[\chi_H,1_H]=\chi(1)$, then $\chi_H=\chi(1)1_H+\sum a_\varphi \varphi$, for certain non-trivial $\varphi \in Irr(H)$ and non-negative integers $a_\varphi$. By comparing degrees left and right in this equation we see that $a_\varphi=0$, for each of these irreducible constituents $\varphi$ of $\chi_H$.

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