Let $\zeta$ be the $63^{th}$ root of unity. The reason why I choose the number $63$ is because $$\operatorname{Gal}(\Bbb Q[\zeta]/\Bbb Q)\simeq (\Bbb Z/63\Bbb Z)^\times\simeq(\Bbb Z/2\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\oplus(\Bbb Z/3\Bbb Z)\oplus(\Bbb Z/3\Bbb Z)$$
One could find generators in $\operatorname{Gal}(\Bbb Q[\zeta]/\Bbb Q)$ corresponding to the two subgroups of order $3$: $$\left<\sigma_4\right>=\{\operatorname{id},\sigma_4,\sigma_{16}\}$$$$\left<\sigma_{25}\right>=\{\operatorname{id},\sigma_{25},\sigma_{58}\}$$
It turns out from the structure of $(\Bbb Z/63\Bbb Z)^\times$ that the subgroup $\left<\sigma_4,\sigma_{25}\right> = \left<\sigma_2^2,\sigma_{5}^2\right>$ of order $9$ is not cyclic. We could not apply the usual way of writing $$\operatorname{Fix}\left<\sigma_k\right>=\Bbb Q[\zeta+\zeta^k+\zeta^{k^2}+...]$$ but $\operatorname{Fix}\left<\sigma_4,\sigma_{25}\right>$ must also be simple over $\Bbb Q$. How could we compute the primitive element in this case?
