The galois group of a polynomial of degree 3 is either $A_3$ or $S_3$

Hungerford -Algebra p.271

Let $E/F$ be a Galois extension where $E$ is a splitting field for a separable irreducuble polynomial $f$ over $F$ whose roots are $a_1,a_2,a_3$.

Let $\Phi:Gal(E/F)\rightarrow S_n$ be the natural group monomorphism permuting roots of $f$.

Then the order of '$Gal(E/F)$ is divisible by $3$, henve it is a normal subgroup of $S_3$.

These are all I know. How do I prove that the Galois group is eithet S3 or A3?

Hint: Why must $\text{Gal}(E/F)$ act transitively on the roots of $f$? What are the transitive subgroups of $S_3$?
$S_3$ has only six elements, all either of order 2 or 3 or the identity (three elements of order 2, two of order 3). A subgroup with order divisible by 3 must by Cauchy's theorem contain an element of order 3, hence it must contain both elements of order 3 (since each is the square of the other).
The subgroup consisting of the identity and the two elements of order 3 is $A_3$. If we add an element of order 2, we end up getting all of the rest of the elements of order 2 since a subgroup with an element of order 2 and an element of order 3 must have at least six elements. But that means it is the entire group!
Thus the only possibilities are $A_3$ and $S_3$.