- |U| - 1).$$ Hence we have $$(**) \ \ \ \ \ \ \sum_{u \in \Pi_2}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$= 1 - (1_{G}^{(1, b(G))}, 1_{G}^{(1, b(G))}) 
-  \sum_{u \in \Pi_1}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$=  \frac{(e - 1)|P| - (e - 1)p^n  + |U|}{e|P|}.$$ Since $(\eta_{\hat{\nu}}, 1_{P}) = 0$, $$e + \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u) e |P :C_{P}(u)| + \sum_{u \in \Pi_2}\eta_{\hat{\nu}}(u) |P : C_{P}(u)| = 0,$$ and hence $$(***) \ \ \ \ \ \ \  \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u) |C_{P}(u)|^{-1}$$ $$= -|P|^{-1} - \sum_{u \in \Pi_2} |C_{P}(u)|^{-1}   = -|P|^{-1} - |P|^{-1}|U|$$ because $\eta_{\hat{\nu}}(u) = e$ for $u \in \Pi_2$. From $(*), (**)$ and $(***)$ $$( 1_G * \eta_{\hat{\nu}}, 1_G)$$ $$= e(1_{G}^{(1, b(G))}, 1_{G}^{(1, b(G))})  +  \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u)(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)}) +  e\sum_{u \in \Pi_2}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$= \frac{(e - 1)p^n + 1}{|P|}  + (- |P|^{-1} - |P|^{-1}|U|) + \frac{(e - 1)|P| - (e - 1)p^n + |U|}{{|P|}}$$ $$= e - 1.$$ From this we have $$(1_G * \eta_{\hat{\nu}}, 1_G * \eta_{\hat{\nu}} ) =  
(1_G * \eta_{\hat{\nu}}\eta_{\hat{\nu}^{-1}}, 
1_G) =  e + (e-1)^2 = e^2 - e + 1.$$ Moreover, for $\mu, \mu' (\neq) \in {\cal M}$, $$(1_G * \eta_{\hat{\mu'}}, 1_G * \eta_{\hat{\mu}} ) =  (1_G * \eta_{\hat{\mu'}}\eta_{\hat{\mu}^{-1}}, 1_G) =  