Exercise 8.10

Proof. ${(\mathit{\chi \rho })}^6={\chi }^6{\rho }^6=𝜀,{(\mathit{\chi \rho })}^2={\chi }^2\ne 𝜀,{(\mathit{\chi \rho })}^3={\rho }^3=\rho \ne 𝜀$, so $\mathit{\chi \rho }$ is of order 6.

$J(\chi ,\rho )g(\mathit{\chi \rho })=g(\chi )g(\rho )$ since $\chi ,\rho ,\mathit{\chi \rho }$ are non trivial characters. So

From Exercise 8.9, $g{(\chi )}^6=p^2{\pi }^2$. Proposition 6.3.2 gives $g{(\rho )}^2={(-1)}^{(p-1)∕2}p$, so $g{(\rho )}^6={(-1)}^{(p-1)∕2}{p}^3$. As $\pi =\chi (2)J(\chi ,\rho )$, $J{(\chi ,\rho )}^6=\chi {(2)}^{-6}{\pi }^6={\pi }^6$, since $\chi {(2)}^3=1$. Therefore

Moreover, $\pi \bar{\pi }=\chi (2)\overline{\chi (2)}J(\chi ,\rho )\overline{J(\chi ,\rho )}=|J(\chi ,\rho ){\text{|}}^2=p$ (Theorem 8.1, Corollary), so ${\pi }^{-1}=\bar{\pi }∕p$. In conclusion,

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