# Subgroup of free products is torsion-free

Let group $G$ be a free product of finitely many finite groups $G_i$: $G=G_1\ast \cdots\ast G_n$.

Suppose that $g_1, g_2\in G$ and elements $g_1,g_2$ have infinite order in $G$ and $[g_1,g_2]\ne 1$.

What conditions must satisfy these elements in order to subgroup $\left \langle g_1,g_2 \right \rangle$ was torsion-free (in other words $\left \langle g_1,g_2 \right \rangle\simeq F_2$)?

• Necessary and sufficient conditions are that $g_1,g_2$ either commute or satisfy no nontrivial relations. What kind of "conditions" are you looking for beyond that? Apr 1, 2016 at 11:27
• @arctictern Maybe you are right and there is nothing better to say. But, for example, if $g_1=a$ and $g_2 = a^b$ then it is not easy to prove, that trere is no such relation. Apr 1, 2016 at 11:34
• What are $a$ and $b$ in your comment? Generators of $F_2$? Apr 1, 2016 at 11:57
• $a$ and $b$ are such elements that $g_1$ and $g_2$ have infinite order and $[g_1,g_2]\ne 1$. Apr 1, 2016 at 12:03

Let $T$ denote the Bass-Serre tree associated to the free product. Therefore, $G$ acts on $T$ such that:
• edge-stabilisers of $T$ are trivial;
• vertex-stabilisers of $T$ are finite.
Now, if $g,h \in G$ have infinite order, they are loxodromic isometries of $T$. Two cases may happen. First, the axes of $g$ and $h$ may have finite intersection. Then, playing ping-pong, it is classifical to prove that $\langle g, h \rangle$ is a free group of rank two. Second, the axes of $g$ and $h$ may have infinite intersection. Then, it is not difficult to find an edge which belongs to this intersection and two exponents $m,n \in \mathbb{Z} \backslash \{0 \}$ such that $g^mh^{-n}$ stabilises this edge. Since edge-stabilisers are trivial, we deduce that $g^m=h^n$.
Consequently, two infinite-order elements $g,h \in G$ generate a free subgroup of rank two if and only if $g^m \neq h^n$ for every $m,n \in \mathbb{Z} \backslash \{ 0\}$.
• Let $G = \left \langle a \right \rangle_2\ast\left \langle b \right \rangle_3$ and $g_1 = ab$ and $g_2 = ba = g_1^a$. It is clear that there is no such $n$ and $m$, that $g_1^n = g_2^m$, but $g_1g_1 = b^2$. Apr 4, 2016 at 17:24
• You're right. In fact, my argument only shows that $\langle g^p,h^q \rangle$ is free for some $p,q \geq 1$. The problem is not as simple as I suspected first. I will delete my answer. Apr 4, 2016 at 20:13
• Actually I believe that it is free if $|p|, |q|\ge 2$. Apr 4, 2016 at 20:31