Q: If $Gal(L/K)$ can be embedded into $S_n$, then must $L/K$ be the splitting field of some polynomial of degree $n$? This year I'm a teaching assistant of an undergraduate Galois theory course. The professor left a bonus question for the students:
Suppose $L/K$ is a finite Galois extension with $G:=Gal(L/K)$ that can be embedded into $S_n$. For convenience, let's identify $G$ with that subgroup: $G\subset S_n$. Question: must $L/K$ be the splitting field of some separable polynomial $f\in K[x]$ of degree $n$ (not necessarily irreducible)? (I found that we must avoid the case of finite fields. So we assume $K$ to be an infinite field.)
I got stuck. If $G=S_n$ or some other special subgroups, I know some ad-hoc methods to treat the problem, such as
Field Extension with Galois group $S_n$ is the splitting field of a polynomial of degree $n$
But now this $f$ may not be irreducible. I think I should first reduce the problem to the case where $G$ is a transitive subgroup of $S_n$. However, so far I can neither finish the reduction process nor prove this while $G$ is a transitive subgroup. My naive idea is as follows:
(supplemented after Balaji's answer) First we take a primitive element $\alpha$ of $L/K$. Let $f(x)\in K[x]$ be the minimal polynomial of $\alpha$. The $L/K$ is the splitting field of this $f(x)$. If $deg(f)\leq n$, then we are done. (We assumed $K$ to be infinite, then we just take something like $f(x)(x-x_1)...(x-x_{n-m})$.) The problem is: how about $deg(f)>n$? It may happen that a transitive subgroup of $S_k$ with $k>n$ can be embedded into $S_n$. In this case what can I do?
Can anyone give me some help/suggestions? Thanks a lot in advance!

A: $L/K$ is Galois, $G=Gal(L/K)$ is a subgroup of the permutations of $1\ldots n$.
Let $Fix(i)=\{g\in G,g(i)=i\}$ and $F_i=L^{Fix(i)}$ the subfield of $L$ fixed by $Fix(i)$.

What is $E_i$ the Galois closure of $F_i/K$ ?

It is $$E_i = L^{\bigcap_{g\in G} gFix(i)g^{-1}}$$
Let $G(i)=\{ g(i),g\in G\}$ and $Fix(G(i))$ the subgroup of $G$ leaving every element of $G(i)$ fixed. Then $E_i$ is the subfield of $L$ fixed by $Fix(G(i))$.

What is $[F_i:K]$ ?

It is $|G(i)|$

Write a disjoint union of orbits $$\{1\ldots n\} = \bigcup_{j=1}^J G(i_j)$$ and let $$F = F_{i_1}\ldots F_{i_J}$$

The Galois closure of $F/K$ is $E_{i_1}\ldots E_{i_J}=L$.
Letting $F_{i_j}\cong K[x]/(u_j(x))$ we get that $L$ is the splitting field of $$u(x)=\prod_{j=1}^J u_j(x)\in K[x]$$
where $$\deg(u)=\sum_{j=1}^J \deg(u_j)=\sum_{j=1}^J|G(i_j)|=n$$

A: Ok, I found that there's already a more precise version of my question, with solution.
Splitting field and polynomial of minimal degree
