Exercise 9.12

Proof. If $\alpha ={(\mathit{\omega \lambda })}^2$, then

So ${\alpha }^2=9\equiv -1(\mathit{mod}5),{\alpha }^4\equiv 1(mod5)$ and ${\alpha }^2\not\equiv 1(\mathit{mod}5)$, thus the class of $\alpha ={(\mathit{\omega \lambda })}^2$ has order 4 in ${(D∕5D)}^{\text{∗}}$, and this implies that $\mathit{\omega \lambda }$ has order 8.

Let $\beta ={\omega }^2\lambda$. $|{(D∕5D)}^{\text{∗}}|=24$, thus ${[\beta ]}^{24}=1$ (where $[\beta ]$ is the class of $\beta$ in $D∕5D$.)

To verify that $[\beta ]$ has order 24, it is sufficient to show that ${[\beta ]}^8\ne 1,{[\beta ]}^{12}\ne 1$.

${\beta }^8={\omega }^{16}{\lambda }^8=\omega {\lambda }^8={(\mathit{\omega \lambda })}^8{\omega }^2\equiv {\omega }^2\not\equiv 1(\mathit{mod}5)$.

${\beta }^{12}={({\omega }^2\lambda )}^{12}={\lambda }^{12}={(\mathit{\omega \lambda })}^{12}\equiv {(\mathit{\omega \lambda })}^4\equiv -1(\mathit{mod}5)$ (since $(\mathit{\omega \lambda })$ has order 8 in $D∕5D$).

Conclusion: $\mathit{\omega \lambda }$ has order 8, $\omega {\lambda }^2$ has order 24 in ${(D∕5D)}^{\text{∗}}$. □