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If we have $[E:F]=n$, where $n$ is not a prime number but is finite, can we like prime factorize $n=p_1p_2...p_r$, so that we have $[E:F]=[E:K_1][K_1:K_2]...[K_{r-1}:F]$ and each of the $[K_i:K_{i+1}]$ has a prime degree of extension? Can we always find such intermediate fields $K_1, K_2,...,K_{r-1}?$

Thank you very much.

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  • $\begingroup$ Trying to mine a counterexample of a group not satisfying the corresponding property for subgroups through groupprops wiki but not much luck. $\endgroup$
    – anon
    Apr 9, 2014 at 5:30
  • $\begingroup$ If this is true, in the cases where $E/F$ is Galois, by Galois Theory, it will imply existence of subgroups of order any given divisor of the Galois group. Is the converse of Lagrange Theorem true for any finite group? (at least for those realizable as Galois groups over $F$). $\endgroup$ Apr 9, 2014 at 5:31
  • $\begingroup$ @PVanchinathan The converse of Lagrange's theorem is false (there is a partial converse for prime power divisors via Sylow theory) but even when the converse of Lagrange's theorem doesn't hold for a group there can still be a sequence $1\le\cdots\le G$ of subgroups each of prime index in the next, so the failure of the converse of Lagrange's is not enough to say this proposition is false. $\endgroup$
    – anon
    Apr 9, 2014 at 5:33
  • $\begingroup$ @user138017 Possibly related: math.stackexchange.com/questions/173828 $\endgroup$
    – Watson
    Aug 16, 2016 at 18:41

2 Answers 2

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In general the answer to your question is negative. For an example let us pick an irreducible quartic polynomial $p(x)$ over the rationals with Galois group $S_4$. If $\alpha$ is one of its roots, then $E=\Bbb{Q}(\alpha)$, $F=\Bbb{Q}$, then $[E:F]=4$, but there are no intermediate fields between $E$ and $F$. For $E$ is the fixed field of a copy of $S_3$, and there are no intermediate subgroups $H$ of order 12 such that $$S_3\le H\le S_4.$$

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    $\begingroup$ Originally the question carried the tag finite-fields, and the answer catered for the possibility that the asker was only interested in finite fields (in which the answer is affirmative). Apparently that tag was there for the common reason of confusing finite fields and finite field extensions. Just today the answer was upvoted, so my attention was drawn to it. I seized the opportunity to edit out the extras. Sorry about the bump. $\endgroup$ Aug 16, 2016 at 10:55
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No, it's not always possible. If it were, then in particular, every Galois field extension would be solvable, which we know to be false. See http://en.wikipedia.org/wiki/Galois_theory#Solvable_groups_and_solution_by_radicals .

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