Let f be an irreducible polynomial of degree $5$ in $\mathbb{Q}[x]$. Suppose that in $\mathbb{C}$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.
My effort: Let $G$ be the Galois group. The complex conjugation map $\mathbb{\sigma} (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $\alpha_1,\alpha_2, \alpha_3$ be the real roots of $f(x)$ and $\beta_1$ and $\beta_2$ be the nonreal roots. Then, $\sigma$ sends $\beta_1$ to $\beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G \cong S_5$.
I know that any such element $\tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $\tau$?
Thanks!
Proof 1(Direct approach)
Theorem Let $p$ be prime. If a transitive subgroup $H$ of $S_p$ contains a transposition $(1 \ 2)$, then $H=S_p$.
Define a relation $\sim$ on $\{1,2,\ldots, p\}$ by $$ a \sim b \ \Longleftrightarrow (a \ b)\in H. $$
This relation is an equivalence relation (reflexive, symmetric, transitive).
Let $[a]_{\sim}$ be the equivalence class containing $a$. Since $(1 \ 2)\in H$, we have $1 \sim 2$.
Since $H$ is transitive, we can prove that there is a bijection between any two equivalence classes.
Then the cardinality of $[1]_{\sim}$ must divide $p$. Since $\{1,2\}\subseteq [1]_{\sim}$, we must have $|[1]_{\sim}|=p$. This tells us that all transpositions containing $1$ are in $H$. Hence, $H=S_p$.
Proof 2(Galois Theory)
If $G$ is Galois group of an irreducible polynomial $f(x)\in\mathbb{Q}[x]$ of degree $p$, then $p$ divides the order of $G$. Also, $G$ is transitive and it contains a transposition. Since $G$ must have an element of order $p$, it contains a $p$-cycle. Hence $G\simeq S_p$.