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Let $N \lhd G$ (with $G$ finite) and let $\chi \in \mathrm{Irr}(G)$ be such that $\chi_N=\theta \in \mathrm{Irr}(N)$. Then the characters $\beta \cdot \chi$ for $\beta \in \mathrm{Irr}(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\theta^G$. This is Corollary 6.17 of Isaacs' character theory of finite groups.
My question is: Let $\chi \in \mathrm{Irr}(G)$ and set $\mathrm{Irr}(G/N)=\{\theta_1 \dots \theta_n\}$. Suppose that the $\{\theta_i \cdot \chi\}_{i=1}^n$ are distinct and irreducible. Can we conclude that $\chi_N$ is irreducible? If yes, why?

(Notation: with $\chi_N$ i indicate the restriction of $\chi$ to $N$ and with $\theta^G$ i indicate the extension of $\theta$ to $G$. With $\beta \cdot \chi$, $\theta_i \cdot \chi$ i indicate the product of those characters)

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  • $\begingroup$ What's $\beta_\chi$ or $\theta_{i\chi}$ ? $\endgroup$ – Max Feb 18 at 20:00
  • $\begingroup$ @Max it is a product of characters: $\beta \cdot \chi$ and $\theta_i \cdot \chi$. I modified my question adding "\cdot". $\endgroup$ – ciccio Feb 18 at 20:04
  • $\begingroup$ Oh ok I thought it was an index, my bad $\endgroup$ – Max Feb 18 at 20:05
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    $\begingroup$ I don't know why, the $\chi$ stays a Little "lower". $\endgroup$ – ciccio Feb 18 at 20:08
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Let us write and identify $Irr(G/N)=\{\beta \in Irr(G): N \subseteq ker(\beta)\}$. (Note that you are using $\theta_i$ in your question, but this is rather confusing, since you are using $\theta$ for an irreducible character of $N$).

Your assumption is that all the $\beta\chi$ are distinct and irreducible.

Let $\chi \in Irr(G)$ and $\theta \in Irr(N)$ be an irreducible constituent of $\chi_N$. Then $\chi_N=e\sum_{i=1}^{t}\theta_i$, where $t=|G:T|$, the index of the intertia subgroup of $\theta$ in $G$ and $e$ is a positive integer. So $\chi(1)=et\theta(1)$. We need that later.

Now let us have a look at the irreducible constituents of $\theta^G$. By Frobenius Reciprocity we have: $[\beta\chi,\theta^G]=[(\beta\chi)_N,\theta]=[\beta(1)\chi_N,\theta]=\beta(1)[\chi_N,\theta]=\beta(1)e.$ This means that all the different $\beta\chi$ appear with multiplicity $\beta(1)e$ as irreducible constituent of $\theta^G$.

Hence $\theta^G(1)=\theta(1)|G:N| \geq \sum_{\beta}\beta(1)e(\beta\chi)(1)=e\chi(1)\sum_\beta\beta(1)^2=e\chi(1)|G:N|=e^2t\theta(1)|G:N|.$ It follows that $e^2t \leq 1$, meaning $e=1=t$. But then $\chi_N=\theta$ and we are done.

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  • $\begingroup$ All clear. Thank You! I did the same reasoning, but i cannot conclude since i couldn't see that inequality. $\endgroup$ – ciccio Feb 19 at 9:36
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    $\begingroup$ You are welcome. Good question and nice problem, never realized that Gallagher's result had a converse! $\endgroup$ – Nicky Hekster Feb 19 at 9:42

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