My attempt: I understand my argument is wrong but I can't understand where am I making a mistake.
Let $G$ act on $S$ non - trivially.
$$\varphi:G \mapsto \mbox{Sym(S)}$$
$$\varphi(g) \mapsto \tau_g$$
Here $\tau_g:S \mapsto S$ is a bijective-map so that $\tau(g) = g.s$
We know that Ker($\varphi$) is a normal subgroup of $G$ that fixes all elements of $S$.
Then $[G:ker(\varphi)] > 1$ [$G$ acts non-trivially on $S$].
So there exists a non-trival normal subgroup of $G$ which is Ker($\varphi$).
I don't think the solution would be this simple.Where am I making a mistake?How do I rectify it?