3
$\begingroup$

Here is the question again:

Let $F$ be a Galois extension of $K$ such that $[F:K] = 27$. Show that there are intermediate fields of order $9$ and $3$.

I am stumped. By the Galois correspondence, the Galois group of the extension has order 27 and it suffices to find subgroups of order $9$ and $3$. I know by Lagrange's Theorem that intermediate fields of order $9$ and $3$ are allowed.

$\endgroup$
3
  • 1
    $\begingroup$ What do you know about group theory? And how can you connect the search for subfields with the search for subgroups in a group of order 27? $\endgroup$
    – Verdruss
    Jul 9, 2017 at 1:02
  • $\begingroup$ @Verdruss Thanks. The use of "order" where I would usually say "degree" confused me. $\endgroup$
    – sharding4
    Jul 9, 2017 at 1:08
  • $\begingroup$ I am using the Galois correspondence. (Edit in question) $\endgroup$
    – edenstar
    Jul 9, 2017 at 1:10

1 Answer 1

Reset to default
4
$\begingroup$

The Galois group of $F/K$ is a $p$-group.

A $p$-group of order $p^n$ has subgroups of order $p^m$ for all $m \le n$. This follows from the Sylow theorems.

Therefore, the Galois group has subgroups of order $3$ and $9$. The fixed fields of these subgroups have degree $9$ and $3$, respectively.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .