I'm trying to solve this exercise: Find the number of intermediate field extensions $\mathbb{Q}(ζ_{630})/M/\mathbb{Q}$ satisfying $[M:Q]=3$.
Since \begin{gather} Gal(\mathbb{Q}(ζ_{630})/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^{\times} \times (\mathbb{Z}/9\mathbb{Z})^{\times} \times (\mathbb{Z}/5\mathbb{Z})^{\times} \times (\mathbb{Z}/7\mathbb{Z})^{\times} \cong \\(\mathbb{Z}/6\mathbb{Z}) \times (\mathbb{Z}/4\mathbb{Z}) \times (\mathbb{Z}/6\mathbb{Z}) \cong G \end{gather} I think the number of such extensions corresponds to the number of subgroups $H$ of $G$ of index $3$. Any subgroup $H$ of index $3$ includes $3G$, so the solution is the number of subgroups of $G/3G \cong (\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/3\mathbb{Z})$ of index $3$. I think those subgroups are generated by $(0,1), (1,0), (1,1), (1,2)$, so the solution is $4$. However, the book's solution is $8$. Where is my mistake?