The result if $E$ is the splitting field of a separable polynomial of degree $n$ over $F$, then $Gal_F(E)$ is isomorphic to a subgroup of $S_n$ is fairly common.
However I am wondering if the converse (plus a few conditions) is also true.
I think it is true but am not sure of how to prove it.
Thank you for any assistance.
$S_3$ contains exactly 3 transpositions, and the subgroups they generate are obviously not normal. The fixed field $K$ of one of them is then a non normal extension of $F$ of degree 3, whose normal closure is $E$. In other words, $E$ will be the splitting field over $F$ of an irreducible cubic $f\in F[X]$, the minimal polynomial of a chosen primitive element of $K$, as hinted by Brandon Carter.
But more can be said in the converse direction, as you probably know. If $f\in F[X]$ an irreducible cubic, its Galois group $G$ is a subgroup of $S_3$ with order divisible by 3, hence $G \cong A_3$ or $S_3$. For char$F\neq 2$, $G \cong A_3$ iff disc ($F$) is a square in $F$.