Given quintic polynomial $f$ - If ${\rm Gal}(f)$ has an element of order 3 then it is not solvable

Question

Let $f(x) \in \mathbb Q[x]$ be a quintic irreducible polynomial, $E$ be the splitting field of $f(x)$ over $\mathbb Q$. Assume ${\rm Gal}(E/\mathbb Q)$ has a subgroup (or an element) of order $3$. Prove $f$ is not solvable by radicals over $\mathbb Q$

I'm trying to show that ${\rm Gal}(E/\mathbb Q)$ is not a solvable group, perhaps using this claim but can't find a way to show it. I have tried to regard ${\rm Gal}(E/\mathbb Q)$ as a subgroup of $S_5$ and then I know that an element of order $3$ is a cycle, say WLOG $(1\space2\space3)$. Can I conclude something from that?

Any help would be appreciated

Answer 1

It is a classic theorem that the maximal solvable transitive subgroup of $S_5$ consists of all transforms

$$n\mapsto an+b$$ where we have arithmetic modulo $5$. Here $0\leq b\leq 4$ and $1\leq a \leq 4$. This group has order $20$. From this the claim follows.

Answer 2

Here's a proof that a $3$-cycle and a $5$-cycle generate $A_5$. Let $G$ be the subgroup of $A_5$ generated by a $3$-cycle $\sigma$ and a $5$-cycle $\tau$. Note that there are $\binom{5}{3}=10$ subgroups of order $3$ in $A_5$. By the Sylow theorems, the normalizer of a $3$-Sylow has index $10$, so it consists of $6$ elements. As we have $N_{G}(\langle \sigma \rangle) \subset N_{A_5}(\langle \sigma \rangle)$, $\tau$ (and in fact any $5$-cycle) does not normalize $\langle \sigma\rangle$. Hence the index $[G:N_{G}(\langle \sigma \rangle)]$ is divisible by $5$. But by Sylow theory, we have $[G:N_{G}(\langle \sigma \rangle)] \equiv 1 \pmod 3$, thus we get that $10$ divides $[G:N_{G}(\langle \sigma \rangle)]$ and hence $|G|$. So we get that $30$ divides $|G|$. But if $|G|=30$, then $[A_5:G]=2$, so $G$ is normal in $A_5$, contadicting the fact that $A_5$ is simple. Thus $G=A_5$, as claimed.

Why does this solve the question? Well, since $f$ is irreducible, we have that $5$ divides $[E:\Bbb Q]$, so $\mathrm{Gal}(E/\Bbb Q)$ contains an element of order $5$, i.e. a $5$-cycle.