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

Question

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)

Answer 1

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 $5$,