I am struggling on the problem related to the free product of groups.
Let $G$ be a group acting on a set $X$ and let $H$ be a group generated by two subgroups $G_1, G_2$ of $G$ with $\lvert G_1 \rvert \geq 3, \lvert G_2 \rvert \geq 2$. Suppose that there exist two nonempty subsets $X_1$ and $X_2$ of $X$ such that $X_2 \not\subset X_1, g(X_2) \subseteq X_1$, and $g’(X_1) \subseteq X_2$ for any nontrivial $g \in G_1$ and $g’ \in G_2$. Show that $H \simeq G_1 \ast G_2$.
I think I have to use the universal property of the free product of groups, but I have no idea on how to apply it. Any comments or hints are welcomed.
To prove that $H = G_1*G_2$, it is enough to prove that any nonempty alternating product of nontrivial elements from $G_1$ and $G_2$ is nontrivial in $G$.
The assumptions $|G_1| \ge 3$ and $|G_2| \ge 2$ imply that any such element of $G$ has a conjugate $g$ of the form $a_0b_1a_1b_2a_2\cdots b_ka_k$ for some $k \ge 0$ with each $a_i \in G_1 \setminus \{1\}$ and $b_i \in G_2 \setminus \{1\}$.
Let $x \in X_2 \setminus X_1$. Then the assumptions imply that $g(x) \in X_1$, so $g(x) \ne x$ and hence $g \ne 1$, and so the original element, a conjugate of $g$, is also nontrivial.