# Show Galois extension with $[K : F] = n$ has an intermediate filed with $L$ with $[K : L] = p$, where $p$ is a prime divisor of $n$

It is a two part question.

Let $K/F$ be a Galois extension with $[K : F] = n$. If $p$ is a prime divisor of $n$, prove there is an intermediate field $L$ with $[K : L] = p$. Prove or disprove that there is an intermediate field $M$ with $[M : F] = p$.

Could anyone help me for some hints please, Thanks a lot!

For the second part, you’d want a subgroup with index $p$,and these will not necessarily exist, as, for instance, when $p=2$ and your Galois group is simple and noncyclic. For, a subgroup of index $2$ has to be normal.
• Thanks for answer, here is what I think from your hint: as $[K:F]=n$, we have $Gal(K/F)=n$, since $p|n$, then by Cauchy's theorem，$Gal(K/F)$ has an element (say $g$) of order $p$, so by Galois' correspondence theorem, there is a intermediate field $L$ fixed by $g$ with $[K:L]=p$? Am I right? Dec 6 '15 at 23:17
• Perfect! :-) Don't forget the cardinality when you write $|Gal(K/F)|=n$. Dec 6 '15 at 23:20