I had some trouble handling the question below:
Let $G$ be a group acting on a set $S$ containing at least two elements. Assume $G$ is transitive and $\{g\in G:gx=x\ \text{for all}\ x\in S\}=\left<e\right>$. If $N\lhd G$ and $N<G_x$ for some $x\in S$, then $N=\left<e\right>$.
Some terminologies are explained here:
(i) We say $G$ is transitive, if given any $x,y\in S$, there exists $g\in G$ such that $gx=y$.
(ii) $G_x$, the stablizer of $x$, which also denoted as $\mathrm{Stab}(x)$, is the set $\{g\in G:gx=x\}$.
Here is my attempt. For any $n\in N$, since $n$ is a subgroup of $G_x$ we have $nx=x$. Then $N$ is normal in $G$ implies $gng^{-1}\in N$ for all $g\in G$. But I can't go any further. Can someone offer me any suggestions? Thanks in advance.
$Nx = x$ for some $x \in S$.
Hence $gNx = Ngx = gx \implies N$ fixes $Gx$.
But $G$ is transitive. Hence $Gx = S$.
Hence $N$ fixes $S$.
Hence $N = \langle e\rangle$.