Let $G$ be a finite group, and $U \le G$ of odd order such that $N_G(U) = TU$ with $T = \langle t \rangle$ for some involution $t$ and assume that $U^g \ne U$ implies $U^g \cap U = 1$. Then I can not understand the following argument:
Let $\hat \lambda : TU \to \mathbb C$ be a linear character on $TU$. Further set $S = TU \setminus T^U$ and suppose $N_G(S) = TU$ and for $g \notin N_G(S)$ we have $S^g \cap S = \emptyset$ and that $\hat \lambda$ is zero outside of $S$ and that $$ ((1-\hat \lambda), (1-\hat\lambda))_{TU} = 2 \quad\mbox{and}\quad (1_{TU}, (1- \hat\lambda))_{TU} = 1. $$ Then there exists some linear character $\Lambda : G \to \mathbb C$ such that $(1-\hat\lambda)^G = 1 - \Lambda$.
I have some difficulty following the last line, I do not comprehend: What is $(1 - \hat \lambda)^G$ and why does it equals $ 1 - \Lambda$ for some linear character on $G$?
So could anyone please explain what is happening here?? Thank you!