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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!

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2 Answers 2

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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.

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  • $\begingroup$ +1 A nice and simple counterexample, I hadn't thought about it! $\endgroup$
    – C. Falcon
    Dec 6, 2015 at 23:38
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Hint. For the first part, use Cauchy's theorem and Galois' correspondence theorem.

Let me know if you need a more detailed answer.

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  • $\begingroup$ 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? $\endgroup$
    – Michael
    Dec 6, 2015 at 23:17
  • $\begingroup$ Perfect! :-) Don't forget the cardinality when you write $|Gal(K/F)|=n$. $\endgroup$
    – C. Falcon
    Dec 6, 2015 at 23:20
  • $\begingroup$ Thank you! could you give me some hint for part two as well? Is it the statement true or false? $\endgroup$
    – Michael
    Dec 6, 2015 at 23:25
  • $\begingroup$ The second statement might be false, but I don't have counterexample right now. Let me think about it. $\endgroup$
    – C. Falcon
    Dec 6, 2015 at 23:26

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