# About a Corollary of Isaacs' “character theory of finite groups”: does the converse implication hold?

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)

• What's $\beta_\chi$ or $\theta_{i\chi}$ ? – Max Feb 18 at 20:00
• @Max it is a product of characters: $\beta \cdot \chi$ and $\theta_i \cdot \chi$. I modified my question adding "\cdot". – ciccio Feb 18 at 20:04
• Oh ok I thought it was an index, my bad – Max Feb 18 at 20:05
• I don't know why, the $\chi$ stays a Little "lower". – ciccio Feb 18 at 20:08

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.