The task is the following:
Show that a Galois extension $L/K$ of degree $45$ has got at most $12$ intermediate field extensions.
Below I present a proof. I seek a more general method for this kind of problems.
Using Galois correspondence, each intermediate field extension corresponds to a subgroup of $Gal(L/K)$ and $|Gal(L/K)|=45$.
Using Sylow theorems, we show that there are one of each Sylow subgroups for $3,5$ and conclude that they are normal subgroups and thus their product generates the whole group.
The Sylow-$5$ group must be isomorphic to $\Bbb{Z}_5$ and the Sylow-$3$ group must be isomorphic to either $\Bbb{Z}_9$ or $\Bbb{Z}_3 \times \Bbb{Z}_3$.
$\Bbb{Z}_5 \times \Bbb{Z}_9 \cong\Bbb{Z}_{45}$ is cyclic so every divisor of $45$ gives exactly one subgroup, total of $6$.
$\Bbb{Z}_5 \times \Bbb{Z}_3 \times \Bbb{Z}_3$ has the subgroups:
- Order $1$: $\langle 1 \rangle$
- Order $3$: $\langle (010) \rangle$, $\langle (001) \rangle$, $\langle (011) \rangle$, $\langle (012) \rangle$
- Order $5$: $\langle (100) \rangle$
- Order $9$: $\langle (010),(001) \rangle$
- Order $15$: $\langle (100), \sigma \rangle$ for each $\sigma$ of order $3$, total of $4$
- Order $45$: whole group
Which gives exactly $12$ subgroups.
However the last part seemed like a proof by exhaustion and it seemed "lucky", the group was pretty small and I could just write out the subgroups.
Are there more elegant methods for this problem? General ones or in the case that the group is a product of normal subgroups? Thanks in advance.