# Galois group acting transitively on the set of roots

Let $$f\in F[x]$$ be an irreducible polynomial of degree 4, and let $$E$$ be a splitting field of $$f$$ over $$F$$. Can $$Gal(E/F)$$ be isomorphic to the symmetric group $$S_3$$.

My guess is that it cannot be, and my approach is to condition on the number of distinct roots. But my difficulty is to argue for the case when the number of distinct roots is 3.

For three distinct roots, I know $$f(x) = (x-r_1)^2(x-r_2)(x-r_3)$$. I would guess that the key is to show that the Galois group cannot act transitively on these sets of roots, but I cannot figure out exactly how to show that. Any hint on this problem?

Let $$a$$ be a root of $$f$$ and consider the extension field $$F(a)$$. Since the minimal polynomial of $$a$$ is of degree four, we get that $$[F(a):F]=4$$. Hence the degree of the splitting field extension is a multiple of four, and thus so is the size of the Galois group. But $$4\nmid |S_3|=6$$, which completes the proof.
• We know from the Fundamental Theorem of Galois Theory that the degree of the splitting field extension is equal to the size of the Galois group. So if $4$ is a divisor of one of those quantities, it must also divide the other. Commented May 13, 2020 at 1:15
• To use the fundamental theorem don't we have to assume that the extension is Galois? If I am interpreting correctly, the question here doesn't assume $f$ to be separable. Commented May 13, 2020 at 3:51
A variation on the answer already given may be as follows. Let $$\alpha_{1},\cdots, \alpha_{4}$$ be the roots of $$f(x)$$. By its operation on the roots, the Galois group represents a transitive subgroup of $$S_{4}$$ whose order by the Counting Formula (or orbit-stabilizer theorem) must be divisible by 4, which rules out $$S_{3}$$.