Let $L/\mathbb{Q}$ be a Galois closed field with Galois group a subgroup of $S_6$. Is it the case that $L$ is the compositum of Galois closures of linearly disjoint fields of degree at most $6$ over $\mathbb{Q}$?

Currently this looks true from a bashing argument: taking each subgroup of $S_6$ up to isomorphism and either noting it is a direct product, or it has a large subgroup which is not normal and whose normalizer is the whole subgroup. Bashing argument isn't complete yet, so this might still be false. I'd like something more elegant though, and if it works for $S_n$ even better.

Galois fieldusually means a finite field, I guess you meanGalois extension. (ii) Apparently you want the Galois group to be isomorphic to a subgroup of $S_6$. (iii) Doesdisjointmean that the intersection is $\mathbb Q$, or does it meanlinearly disjointover $\mathbb Q$ which is somewhat stronger? $\endgroup$ – Peter Mueller May 2 '17 at 13:52