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

Your stabilizer is not complete, it is not a subgroup, for example!
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.  
