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 $5$,


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.