Talk:Aufgaben:Problem 14

What about

Define \(Z_{\hat{G}}:=\{g\in G:\chi(g)=1\; \forall \chi \in \hat{G}\}\)

Lemma 5' : \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x \in Z_{\hat{G}}} \)

Proof of Lemma 5': let \(\rho \in \hat{G}\) be arbitrary

$$\sum\limits_{\chi}{\chi(x)}=\sum\limits_{\sigma=\rho^{-1}\chi}{\sigma(x)\rho(x)}=\rho(x)\sum\limits_{\sigma \in \hat{G}}{\sigma(x)}$$ $$\Rightarrow \rho(x)=1 \;\forall \rho \in \hat{G}\; or \; \sum\limits_{\chi}{\chi(x)}=0$$ $$\Rightarrow \sum\limits_{\chi}{\chi(x)}=\delta_{x \in Z_{\hat{G}}}\sum\limits_{\chi}{1} =|G|\delta_{x \in Z_{\hat{G}}}$$

Lemma 6: \(Z_{\hat{G}}=\{e\}\)

Proof of Lemma 6: Use Lemma 1

$$ \sum\limits_{\chi}{\sum\limits_{g}{\chi(g)}}=\sum\limits_{\chi}{|G|\delta_{\chi, \tau}}=|G|$$ $$ \sum\limits_{g}{\sum\limits_{\chi}{\chi(g)}}=\sum\limits_{g}{|G|\delta_{g \in Z_{\hat{G}}}}=|G| \cdot |Z_{\hat{G}}| $$

\( \Rightarrow |Z_{\hat{G}}|=1 \), and as \(e \in Z_{\hat{G}} \Rightarrow Z_{\hat{G}}=\{e\}\)

Therefore It's proven that \(\forall \chi \in \hat{G}, \; \chi(g)=1 \Leftrightarrow g=e\) and Lemma 5' becomes Lemma 5: \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x,e} \)

So you can do the whole exercise without proving first that \(\{\chi\}\) is a basis of \(L(G)\).