Let G be a group of order n that operates nontrivially on a set of order r. Prove that if n > r!, then G has a proper normal subgroup.
Also I am not very clear of the term "operates nontrivially", does trivially means for any element g in G and s in S, there is gs = s?
When you have a group action and are asked to find a normal group that isn't $G$ or $\{1\}$, it usually helps to look at the kernel of the permutation representation that corresponds to the action.
In this problem we have $\varphi : G \to S_r$, where $|S_r|=r!$.
Since $|G|=n>r!=|S_r|$, the function $\varphi$ cannot be injective, meaning $\ker\varphi \neq \{1\}$.
Now all we need to show is that the kernel is not all of $G$ either.
This is where we use the hypothesis that $G$ acts nontrivially on the set; this tells us that there exists a $g \in G$ such that $gs \neq s$. Equivalently, this $g$ is not in the kernel, which means that $\ker\varphi\neq G$, so it is a proper normal subgroup of $G$.