Let $\Omega/\mathbb{Q}$ be the splitting field of an irreducible polynomial $f \in Q[X]$ of degree five. Show that $Gal(\Omega/\mathbb{Q})$ equals $A_5$ or $S_5$ if $Gal(\Omega/\mathbb{Q})$ has an element of order three.
What I tried
I write $G = Gal(\Omega/\mathbb{Q})$. It is enough to show that the $|G| \geq \frac{1}{2}\cdot 5 \cdot 4 \cdot 3 \cdot 2 = 60$. Let $\sigma$ be the element with order three. We can write $\sigma$ as a product of disjoint cycles, and since $3$ is prime, it must have the form $(abc)$. The polynomial odd amount of reals roots. So.... any hints?
Since $f$ is irreducible, $G$ must act transitively on the roots. Thus the stabilizer of a point is a subgroup of index $5$, so $|G|$ is divisible by $5$. Since $|G|$ is also divisible by $3$, it must be either $15,30$ or $60$. If $|G|=15=3\cdot 5$ then $G$ must contain an element of order $5$ by Cauchy's theorem, so must contain a $3$-cycle and a $5$-cycle, and these must generate the group. If there were $6$ subgroups of order $5$ they would have at least $6\times 4 + 1=25$ elments, so there is only $1$ which is normal by Sylow's theorems. Thus $G\cong \mathbb Z_5\rtimes \mathbb Z_3$. But there is no nontrivial map $\mathbb Z_3\to \mathrm{Aut}(\mathbb Z_5)\cong \mathbb Z_5^\times$, so this semidirect product must be direct, hence $G$ is abelian. But no $3$-cycle and $5$-cycle commute, a contradiction. Hence $|G|$ is either $30$ or $60$.