It is known that $\varphi:G\times X\to X$ is a group action of group $G$ on non-empty set $X$ if
$\varphi(e, x)=x$,
$\varphi(hg, x)=\varphi(h, \varphi(g, x))$ for all $g, h\in G$ and for all $x\in X$.
Assume that $G$ is finitely generated group and generating set is $S$ i.e. $G=\langle S\rangle$.
It is clear that if $\varphi:G\times X\to X$ is a group action, then $\varphi(s_1s_2, x)=\varphi(s_1, \varphi(s_2, x))$ for all $s_1, s_2\in S$ and for all $x\in X$ and $\varphi(e, x)=x$.
What can say about the converse of it? Is it true that if
- $\varphi(e, x)=x$, $\forall x\in X$,
- $\varphi(s_1s_2, x)=\varphi(s_1, \varphi(s_2, x))$ $\forall s_1, s_2\in S$ and all $x\in X$
then $\varphi:G\times X\to X$ is a group action?
Please help me to know it.
Take any cyclic group $G = \langle s\rangle$ with $s$ of order $n \geq 3$, $X = \{e,s\}$, and let: $$\varphi: (g,x) \in G \times X \mapsto \begin{cases}x \quad\text{if } g \neq s^3 \\ s \quad\text{if } g = s^3\end{cases}$$ Then it satisfies: $\varphi(e,x) = x$ and $\varphi(s^2,x) = x = \varphi(s,\varphi(s,x))$, thus $1$ and $2$ are good, but $\varphi$ is not an group action because: $\varphi(s^3,e) = s \neq e = \varphi(s^2,\varphi(s,e))$ yet $s^3 = s^2 s$.