Showing that the galois group for x^n+ax+b is doubly transitive I read here that if $\alpha$ is a root of $x^n+ax+b $ and $\frac{x^n-\alpha^n}{x-\alpha}=a$ is irreducible, then the Galois group $G$ is doubly transitive.
I don't think I understand the reasoning here. Any hints?
 A: Using the notation of the paper. Remark that 
$$\frac{X^n-\alpha^n}{X-\alpha}-a=\frac{X^n-ax+b}{X-\alpha} $$
When it's written that the equation $\frac{X^n-\alpha^n}{X-\alpha}=a$ is irreducible, it certainly means that $\frac{X^n-\alpha^n}{X-\alpha}-a$ is irreducible.
Let $F:=k(a,b)$ the base field, $K:=k(a,b,\alpha_1,\dots,\alpha_n)$ and $L:=k(a,b,\alpha)$. The Galois group  of $K/F$ is $G$. 
Take $(r_1,r_2)$ and $(s_1,s_2)$ with $r_1\neq r_2$ and $s_1\neq s_2$ be two roots of $X^n-aX+b$. We are looking for $g\in G$ such that $g\cdot r_1=s_1$ and $g\cdot r_2=s_2$.


*

*Justify and use the fact that the action of $G$ on $\{\alpha_1,\dots,\alpha_n\}$ is transitive to reduce to the case where $r_1=s_1=\alpha$. 

*With the reduction above, we are looking for $g\in G$ fixing $\alpha$ and sending $r_2$ to $s_2$. Justify that the Galois group of $K/L$ acts transitively on $\{\alpha_1,\dots,\alpha_n\}-\{\alpha\}$ and conclude.
Note : I am not seeing exactly why the polynomials are irreducible...
In general, If a polynomial $P(X)$ is irreducible over $k$. Then its Galois group over $k$ acts transitively on the set $R$ of roots of $P$. If, furthermore for some root $r\in R$ $P(X)/(X-r)$ is also irreducible over $k(r)$ then its Galois group acts doubly transively on the set $R$ (using the same method as above). 
