# $G_{x_1x_2+x_3x_4}\cong D_8$

Let $$R=F[x_1,x_2,x_3,x_4]$$ be the set of polynoms in 4 variables over a field $$F$$. Let a map $$\varphi:S_4\to \operatorname{Sym}(R)$$ by $$\\ (f(x_1,x_2,x_3,x_4))\varphi(\sigma)=f(x_{1\sigma},x_{2\sigma},x_{3\sigma},x_{4\sigma}). \$$ Prove that $$G_{x_1x_2+x_3x_4}\cong D_8$$ ($$D_8$$ is the dihedral group).

I've shown that $$\varphi$$ is a group action of $$S_4$$ on $$R$$. My problem is that I know that $$\\ G_{x_1 x_2+x_3 x_4 }=\{σ∈S_4:(x_1 x_2+x_3 x_4 )σ=x_1 x_2+x_3 x_4 \} \ =\{id,(1 2),(3 4),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)\} \$$ then $$|G_{x_1 x_2+x_3 x_4 }|=6\ne 8=|D_8|$$ so I don't see how it is possible that $$G_{x_1x_2+x_3x_4}\cong D_8$$.

To compute the size of the stabilizer $$G$$ of $$p = x_1x_2+x_3x_4$$ you can begin by computing its orbit: it consists of three elements: $$x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $$24/3=8$$.
You already noted that $$G$$ contains $$\{1,(12),(34),(13)(24),(14)(23),(12)(13)\}$$. But this is not a subgroup. It is missing $$(1423)$$, which is $$(13)(24)(12)$$, and it is missing $$(1324)$$, its inverse. This makes $$8$$ elements, so $$G = \{1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)\}$$ is your stabilizer.