# Is there always an intermediate field between non-prime field extension?

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.

• Trying to mine a counterexample of a group not satisfying the corresponding property for subgroups through groupprops wiki but not much luck. – anon Apr 9 '14 at 5:30
• 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 $F$). – P Vanchinathan Apr 9 '14 at 5:31
• @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. – anon Apr 9 '14 at 5:33
• @user138017 Possibly related: math.stackexchange.com/questions/173828 – Watson Aug 16 '16 at 18:41

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 $$S_3\le H\le S_4.$$