If a polynomial has a real and a complex roots, its Galois group is non-abelian. 
Let $f\in\mathbb{Q}[X]$ be an irreducible polynomial. Show, that its Galois group is non-abelian if $f$ has roots in $\mathbb{R}$ and $\mathbb{C}\backslash\mathbb{R}.$

First of all, the separability is not assumed. Is Galois group defined in a same way for this case, i.e. the group of automorphisms of the splitting field of $f$? Supposing it is, define $$\alpha_1 = a +br,~\alpha_2 = c+di; ~a,b,c,d\in\mathbb{Q}, r\in\mathbb{R\backslash\mathbb{Q}}$$ the real and complex roots. Now $d\neq0$ by assumption and $b\neq 0$ because f is irreducible, so $$\alpha_3 = a - br,~\alpha_4 = c-di$$ are also roots, because coefficients of f are rational. Now we have two pairs of conjugate roots, and an argument similiar to one from the answer to the question here can be presumably used:
Non-abelian Galois group

"Let the two nonreal roots be $r,s$, let a real root be $t$. Write $\sigma$ for complex conjugation: then $\sigma(r)=s$, $\sigma(s)=r$, $\sigma(t)=t$. Now Galois groups are transitive, so there's an automorphism $\tau$ with $\tau(r)=t$. Then $\sigma(\tau(r))=\sigma(t)=t$, while $\tau(\sigma(r))=\tau(s)\ne t$."

My question is, how can it be proved that the automorphisms that permute the roots in such fashion are really automorphisms? 
Update: as was pointed out in the comments, irreducible polynomial over a field with characteristic $0$ is separable, hence the splitting field $F$ of $f$ is a Galois extension and $$Gal(F/\mathbb{Q}) = [F:\mathbb{Q}]\geq 4,$$ because $$[F:\mathbb{Q}]\geq[\mathbb{Q(\alpha_1,\alpha_2)}:\mathbb{Q}] = [\mathbb{Q(\alpha_1,\alpha_2)}:\mathbb{Q(\alpha_1)}]\cdot[\mathbb{Q(\alpha_1)}:\mathbb{Q}]\geq2\cdot 2 = 4.$$ So there are at least three automorphisms apart from identity.
Update 2: It seems like I formulated my question vaguely, so I'll try it again. Suppose $deg(f) = n, |Gal(F/\mathbb{Q})| = m$. The only method  I know so far to find all permutations of the roots that define $\mathbb{Q}$-automorphism on $F$ is to eliminate $k = n! - m$ permutations that are not homomorphic. Is there a way to check a specific permutation for being continuable to $\mathbb{Q}$-automorphism? In the problem described above, neither $n$ nor $m$ can be found. Can it be proved, that $\sigma,\tau$ from the cited answer, described only by their action on $2$ and $1$ roots correspondingly, are indeed homomorphisms (and by argument from @rts automorphisms)?
 A: 
My question is, how can it be proved that the automorphisms that permute the roots in such fashion are really automorphisms? 

It is part of the main result of Galois theory that the Galois group of $f$ acts transitively on the set of roots. Hence there exists a $\tau\in G$ that maps the given complex root $r$ to the given real root $t$. On the other hand, complex conjugation $\sigma$ is an automorphism of $\Bbb C$ (and hence also of $\overline{\Bbb Q}$ and of the splitting filed of $f$) and leaves $t$ fixed while mapping $r$ to a different root $s$.
In general, you cannot prescribe more than one value of the permutation of roots (or: The action of $G$ on the thes of roots is not necessarily $2$-transitive). But for the arfument at hand, this is not needed. Observing that $\sigma(\tau(t))=\sigma(r)=s$ and $\tau(\sigma(t))=\tau(t)=r$ is enough.
A: The short answer: $F\mid \mathbb{Q}$ is a normal extension and thus $\sigma(F)=F$. That means the restriction $\sigma|_F$ defines in fact an automorphism of $F$.
Here is a sketch of how to prove this. Let $S$ be the set of roots of $f$.
Then $F=\mathbb{Q}(S)$. Let $\rho: F\to \mathbb{C}$ be any homomorphism. Since $F$ is a field, $\rho$ is injective. Moreover $\rho(\alpha)\in S$ for each $\alpha \in S$,  and thus $\rho$ induces a bijection on the finite set $S$. 
Hence $\rho(F)=\rho(\mathbb{Q}(S))=\mathbb{Q}(\rho(S))=\mathbb{Q}(S)=F$.
