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Consider the following theorem of Brauer (Ref. Isaacs' Character theory, p. 70).

Let $\chi$ be a complex character of a finite group $G$ such that $\langle \chi, 1_G\rangle=0$ (i.e. the trivial character of $G$ is not a constituent of $\chi$). If $A,B\subseteq G$ such that $$\langle \chi_A, 1_A\rangle + \langle \chi_B,1_B\rangle>\langle \chi_{A\cap B}, 1_{A\cap B}\rangle$$ then $A\cup B$ generate proper subgroup of $G$.

Here, the inner product is given by $$\langle \chi_A,1_A\rangle = \frac{1}{|A|} \sum_{a\in A}\chi(a)1_A(a^{-1}).$$ Q.0 The $A$ and $B$ are subsets, not necessarily subgroups, is this right?

Q.1 If $A\cap B$ is empty what should we understand by $\chi_{A\cap B}$? Is it $0$ or $1$? or no such consideration is there on $A\cap B$?

Q.3 What is the (name of) original paper of Brauer for this result?

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  • $\begingroup$ Thanks Jyrki for editing; I wrongly typed the name. $\endgroup$ – Beginner May 18 '18 at 5:47
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    $\begingroup$ I think $A$ and $B$ are subgroups. I think you'll get a helpful example of what happens if we allow subsets by looking at the sign character on $S_3$ and taking $A=\{1,(12)\}$ and $B=\{1,(12), (123)\}$. $\endgroup$ – ancientmathematician May 18 '18 at 6:54
  • $\begingroup$ May be you are right; I asked this Q.0 because of presence of both the notations $A\subseteq G$ and $H<G$ in the referred book. $\endgroup$ – Beginner May 18 '18 at 9:23
  • $\begingroup$ Yes I agree. But even Homer nods. $\endgroup$ – ancientmathematician May 18 '18 at 9:26

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