v_{s+1}$. Suppose $\varepsilon_s=+1$ for now, so that the first letter of $v_{s+1}$ is $\tau^{\pm1}$ by Claim \[pass gamma\]. Then, we have $v_{s+1}(y) \in U_\tau$ by the classical ping-pong argument, and we obtain $y':=\gamma_{i_s}v_{s+1}(y) \in \gamma_{i_s}(U_\tau)$. Now, also by Claim \[pass gamma\], the last letter of $v_s$ is $\sigma^{\pm1}$. Thus, applying the induction hypothesis to $w':=v_1 \gamma_{i_1}^{\varepsilon_1} v_2 \gamma_{i_2}^{\varepsilon_2} \cdots \gamma_{i_{s-1}}^{\varepsilon_{s-1}} v_s$ and $y'$, we see that $w(y)=w'(y') \in U_\sigma \cup U_\tau$. The $\varepsilon_s=-1$ case is analogous.

Since $x \not \in U_\sigma \cup U_\tau$ by the assumption and $w(x) \in U_\sigma \cup U_\tau$ by Claim \[nontrivial\], it follows that $w$ cannot represent the identity element in $\Gamma$. Namely, any non-empty word in $\hat{\calG}$ that represents the identity element of $\Gamma$ must be of length at least $M$ with respect to $\hat{\calG}$. Hence, $U(\Gamma) \geq U(\Gamma; \hat{\calG}\,) \geq M$.

Although every group that satisfies our Infinite Girth Criterion (Proposition \[igc\]) must contain a non-abelian free subgroup, generally groups containing non-abelian free subgroup need *not* have infinite girth. The groups constructed by Olshanskii [@Olshanskii:RelationsBook], which we mentioned in §\[Criteria: Finite Girth\], contain non-abelian free subgroups and have finite girth; indeed, the existence of such non-abelian free subgroups is precisely the reason why those 