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\newtheorem{case}[theorem]{Case}
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\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
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\begin{document}


\section{Serre, Linear Representations of Finite Groups.}

\subsection{Chapter 7}

\QTP{Body Math}
7.3. \ (a) Note that $N_{G}(H)=\{tHt^{-1}=H\mid t\in H\}=H$ by $H\cap
tHt^{-1}=\{1\}$ for every $t\notin H$. Hence there are $[G:H]$ distinct
conjugates of $H$, and $card(\cup _{t\in H}tHt^{-1})=[G:H](card(H)-1)+1=%
\frac{g}{n}(n-1)+1$. Therefore%
\[
card(N)=card(G-\cup _{t\in H}tHt^{-1})=card(G)-card(\cup _{t\in
H}tHt^{-1})=g-(\frac{g}{n}(n-1)+1)=\frac{g}{h}-1.
\]

\QTP{Body Math}
(b) Let%
\[
\widetilde{f}=Ind_{H}^{G}f-f(1)\psi
\]

\QTP{Body Math}
with $\psi $ is the character $Ind_{H}^{G}1-1$ of $G$. Since every class
function on $G$ is the combination of all irreducible characters on $G$, the
uniqueness is trivial, and (c) holds particularly. Pick $s\in H$. $%
\widetilde{f}(s)=\frac{1}{h}\sum_{t\in G,t^{-1}st\in H}f(t^{-1}st)-f(1)(%
\frac{1}{h}\sum_{t\in G,t^{-1}st\in H}1(t^{-1}st)-1)=\frac{1}{h}\sum_{t\in
H}f(s)-f(1)(\frac{1}{h}\sum_{t\in H}1-1)=f(s)$. Hence $\widetilde{f}$
extends $f$. Pick $s\in N$. $Ind_{H}^{G}f(s)=\frac{1}{h}\sum_{t\in
G,t^{-1}st\in H}f(t^{-1}st)=0$, and $Ind_{H}^{G}1(s)=\frac{1}{h}\sum_{t\in
G,t^{-1}st\in H}1(t^{-1}st)=0$ also. Hence $\widetilde{f}(s)=f(1)$. Hence $%
\widetilde{f}=f(1)$ on $N$.

\QTP{Body Math}
(c) By (b).

\QTP{Body Math}
(d) By Frobenius reciprocity, $\left\langle f_{1},f_{2}\right\rangle
_{H}=\left\langle f_{1},\limfunc{Res}\widetilde{f_{2}}\right\rangle
_{H}=\left\langle Indf_{1},\widetilde{f_{2}}\right\rangle _{G}$. Hence we
need to prove $\left\langle \psi ,\widetilde{f_{2}}\right\rangle _{G}=0$, or
$\left\langle \psi ,\widetilde{f_{2}}\right\rangle _{G}=\left\langle
Ind_{H}^{G}1-1,\widetilde{f_{2}}\right\rangle _{G}=0$.
\[
\left\langle Ind_{H}^{G}1,\widetilde{f_{2}}\right\rangle _{G}=\frac{1}{g}%
\sum_{t\in G}Ind_{H}^{G}1(s^{-1})\widetilde{f_{2}}(s)
\]

\end{document}