Exercise 13.3.12

Proof. Suppose $f=x^4-c_1{x}^3+c_2{x}^2-c_{3}x+c_4\in F[x]$ is separable and irreducible. Let $H=⟨(1324),(12)⟩$, $G=⟨(1324))⟩\subset H$ and $\phi =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4)$. By Exercise 4, $G$ is the symmetry group of $\phi$. Since $(12)\cdot \phi =-\phi$, $H\cdot \phi =\{\phi ,-\phi \}$ is the orbit of $\phi$ under the action of $H$. Then

where ${\phi }^2({\alpha }_1,...,{\alpha }_4)=\mathrm{\Delta }(f)(4{\beta }_1+c_1^2-4c_2)$, ${\beta }_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$. By the assumption of Example 13.3.4, ${\beta }_1\in F$, therefore ${\Theta }_f^H(y)\in F[y]$ and it is the relative resolvent in the sense of Exercise 5.

Since $G_f\subset H$, depending on whether ${\Theta }_f^H(y)$ is reducible or not in $F$, the conclusion about whether $G_f=⟨(1324))⟩$ or $G_f=⟨(1324),(12)⟩$ can be done. □