# Degree required to generate a particular galois group

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.

• Maybe your question gets more attention (and correct answers) if it is formulated more clearly. (i) A Galois field usually means a finite field, I guess you mean Galois extension. (ii) Apparently you want the Galois group to be isomorphic to a subgroup of $S_6$. (iii) Does disjoint mean that the intersection is $\mathbb Q$, or does it mean linearly disjoint over $\mathbb Q$ which is somewhat stronger? – Peter Mueller May 2 '17 at 13:52

If you take $E$ be the fixed field of the stabilizer of a point, then $[E:Q]=6$ and $L$ is the Galois closure of $E$. So $L$ is the compositum of the Galois closure of one field of degree $6$.
• What if my subgroup doesn't include the stabilizer of a point? Yes, for $S_6$ itself this is true. – Watson Ladd Apr 30 '17 at 20:26