Suppose $G$ = $ A*_{C}B$, where $|A:C|$ $\geq$ $3$ and $|B:C|$ $\geq$ $2$. How might I show that $G$ has a free subgroup of rank $2$? Here I am not assuming $C$ to be finitely generated.
For me, a reduced word of the amalgam is of the form $(h, s_1, ..., s_n)$ where $h \in C$, and the $s_i$s alternate between $A_1$ and $B_1$ where these are right coset representatives of $C$ in $A$, $B$ respectively and are both assumed to contain $1$ and we have $s_i \ne1$ for all $i$. A cyclically reduced word is such that also $(s_1, s_n)$ is reduced. So if the indexes are as above, there exist distinct cyclically reduced words, and cyclically reduced words have infinite order..
I also know that if $F$ is a subgroup of the amalgam that intersects trivially any conjugate of $A$ or $B$ then $F$ must be free, but I'm not sure how I would apply this.
Hints as to how to proceed with this question (or for more general questions of the same form) would be appreciated.
Thank you.