# Group action on a set that has smaller cardinality than group's one

I have a question that maybe has a trivial answer which I couldn't find though.

Let's consider an action of group $$\mathbf G$$ on a set $$\mathsf X$$ (i.e. homomorphism $$\psi :\mathbf G \to S(\mathsf X)$$ )

So, if I get it right, when $$| \mathbf G | \le |\mathsf X|$$ a non-trivial action always exists, because of the Cayley's theorem : $$\mathbf G \overset{\psi}{\to} \mathbf H \subseteq S(\mathbf G)$$ and, since $$| \mathbf G | \le |\mathsf X|$$, $$S(\mathbf G) \overset{id}{\hookrightarrow} S(\mathsf X)$$, so a non-trivial action is $$id \cdot \psi$$

But what if $$| \mathbf G | \ge |\mathsf X|$$? Can it happen that the only possible homomorphism $$\psi :\mathbf G \to S(\mathsf X)$$ is trivial? Can you give an example of such $$\mathbf G$$ and $$\mathsf X$$? Or, vice versa, a non-trivial homomorphism always exists in this case too?

(p.s. I'm just starting to use TeX so I could write symbols $$id$$ and $$\psi$$ right above arrows and I would be very glad if you guys tell me how to do it)

Sure, it can happen. Consider a simple group $$G$$ and a set $$X$$ such that $$|X|!<|G|$$. Then the only homomorphic image of $$G$$ in $$S_X$$ is the trivial group. For example, let $$G=A_5$$ and let $$X$$ have four elements.
Of course there are also examples with $$|X|<|G|\le |X|!$$. For example, let $$X$$ again have four elements, and let $$G$$ be cyclic of order $$5$$,