However, according to the work of Akhmedov in [@Akhmedov:Girth1 §2], the answer turns out to be negative. We also note his observation in [@Akhmedov:Girth1 §4] that certain groups constructed by Olshanskii in his book [@Olshanskii:RelationsBook] have finite girth but do not satisfy any law. Characterizing groups with finite girth appears to be a delicate and difficult task. See [@Akhmedov:Girth1 §5] for related questions.

Criteria for Infinite Girth {#Infinite Girth}
---------------------------

The proof of the Tits alternative for linear groups and mapping class groups, as well as for other classes of groups, use variations of *ping-pong* lemma [@Klein:Ping-Pong] to construct a free subgroup. The following formulation was given in [@Tits:FreeSubgrp].

\[fsc\] Let $\Gamma$ be a group acting on a set $X$. Suppose there exist elements $\sigma, \tau \in \Gamma$, subsets $U_\sigma, U_\tau \subset X$, and a point $x \in X$, such that

1.  $\displaystyle{x \not \in U_\sigma \cup U_\tau}$,

2.  $\displaystyle{\sigma^k ( \{x\} \cup U_\tau ) \subset U_\sigma }$ for all $k \in \Z-\{0\}$, and

3.  $\displaystyle{\tau^k ( \{x\} \cup U_\sigma) \subset U_\tau }$ for all $k \in \Z-\{0\}$.

Then, $\langle \sigma, \tau \rangle$ is a non-abelian free subgroup of $\Gamma$.

For the proof of the above criterion, one merely observes 