Exercise: Let $G$ be a group generated by a family of subgroup $\{G_i\}_i$. Suppose $G$ operates (acts) on $S$. For each $i$, suppose given subset $S_i\subset S$, and let $s$ be in $S-\cup_i S_i$. Assume that for each $g\in G_i-\{e\}$ $$gS_j \subset S_i \mbox{ for } j\neq i, \hskip1cm g(s)\in S_i.$$ Prove that $G$ is co-product of the family $\{G_i\}_i.$
My attempt: Consider an element $g_1g_2\cdots g_m$. We can assume that $g_i$ and $g_{i+1}$ do not belong to same group in the family $\{G_i\}_i$ (otherwise, treating their product equal to $g_i'$, we can reduce the given element to desired form -no $g_i$ and $g_{i+1}$ belong to same $G_k$).
If $g_1g_2\cdots g_m=1$ (identity) with $m\geq 2$ then $g_1g_2\cdots g_m(s)=s$.
Now $g_m(s)\in S_m$, and $m\neq m-1$, so $g_{m-1}g_m(s)\in S_{m-1}$, continuing this we get $g_1g_2\cdots g_m(s)\in S_1$, and so $s\in S_1$, contradiction.
Question 1. Is this solution correct?
Question 2. For $G_1=\langle x\rangle$ and $G_2=\langle y\rangle$, infinite cyclic groups, can one give example of set $S$ on which a group generated $G_1$ and $G_2$ operates satisfying conditions in the exercise? In other words, can one describe the situation of the exercise by a simple example?