Does this object form a group? Constructing a group knowing a finite set action first. Suppose we have a finite set $S$ and we have defined a set $G$ of maps on $S$, called the generators, with the following properties:


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*For all $g \in G$, $s \in S$, there is $n \in \Bbb{N}, \ n \geq 1$ such that $g^n (s) = s$.

*There exists $s_0 \in S$ such that for all $s \in S$, there is a finite sequence of maps $g_1, \cdots, g_n$ such that $g_1\circ g_2 \circ \cdots \circ g_n(s_0) = s$.
Then does the element-wise composition of $G \circ G \circ \cdots$ converge to a finite group $H$ and also give us a group action $H \times S \to S$ ?
I would say yes.  But am unsure of the proof of it.
 A: No.  For instance, let $S=\{1,2\}$ and let $G=\{g\}$, where $g:S\to S$ is the map that swaps $2$ and $1$.  It is easy to see this satisfies your criteria, but $G^{\circ n}=G$ for $n$ odd and $G^{\circ n}=\{id_S\}$ for $n$ even, so the iterated composites of $G$ with itself do not converge.
By the way, there is a simpler way you express your two conditions.  Condition (1) implies that  for any $g\in G$ and $s\in S$, $s$ is in the image of $g$.  Thus each $g\in G$ is surjective, which means each $g\in G$ is a bijection since $S$ is a finite set.  Conversely, if each $g\in G$ is a bijection, then $g^n=id$ for some $n>0$ since $S$ is finite, and so $g^n(s)=s$ for all $s$.  So condition (1) is equivalent to every element of $G$ being a bijection.
Condition (2) then just says that the group that $G$ generates acts transitively on $S$.  Indeed, notice that since each element of $G$ has finite order, the group generated by $G$ is just the set of compositions of elements of $G$ (you don't need to separately include inverses).  Condition (2) then says there is some $s_0$ whose orbit contains all of $S$, so there is only one orbit.
A: See Burnside Problem. Golod-Shafarevich's negative answer to the General Burnside Problem means that there is an infinite, finitely generated group $G$ in which every element has finite order. Then consider $S=G$ and let $G$ act on itself in the obvious way. Your conditions are satisfied but the generated group is not finite.
