Solution
= Solution
Let $H=\operatorname{Stab}_F(x)$. The identity belongs to $H$ because $F(e,x)=x$. If $g,h\in H$, the <group action> law gives
$$
F(gh,x)=F(g,F(h,x))=F(g,x)=x,
$$
so $gh\in H$. Finally, if $g\in H$, then
$$
x=F(e,x)=F(g^{-1}g,x)=F(g^{-1},F(g,x))=F(g^{-1},x),
$$
so $g^{-1}\in H$. The <subgroup> criterion therefore proves that $\operatorname{Stab}_F(x)$ is the <stabilizer subgroup> of $x$.