$x \in X$ in Proposition \[igc\] satisfy the conditions in Proposition \[fsc\]. Hence, $\langle \sigma, \tau \rangle < \Gamma$ must be a non-abelian free subgroup.

Let $M$ be a positive integer, and we aim to find a new generating set $\hat{\calG}$ for $\Gamma$ such that $U(\Gamma, \hat{\calG}\, ) \geq M$. Let $P=\{ p_1, \cdots , p_n \}$ be a set of positive intergers such that $p_i>M$ for all $i$ and $|p_i-p_j|>M$ for all distinct $i, j$. Let $\hat \gamma_i:= \sigma^{p_i}\gamma_i \tau^{-p_i}$ for each $i$. The set $\hat{\calG}:=\{\sigma, \tau, \hat \gamma_1, \cdots , \hat \gamma_n \}$ clearly generates $\Gamma$. Let $w$ be a nontrivial reduced word in $\hat{\calG} \cup \hat{\calG}^{-1}$ with the length less than $M$ with respect to $\hat{\calG} \cup \hat{\calG}^{-1}$. We can write $w$ as $$w=u_1 \hat \gamma_{i_1}^{\varepsilon_1} u_2 \hat \gamma_{i_2}^{\varepsilon_2} \cdots \hat \gamma_{i_s}^{\varepsilon_s} u_{s+1}\\$$ where $\varepsilon_\ell \in \{ \pm1 \}$ and the subword $u_\ell=u_\ell(\sigma, \tau)$ is a (possibly empty) reduced word in $\{ \sigma^{\pm 1}, \tau^{\pm 1} \}$. If $u_\ell$ is an empty word and $i_{\ell-1} = i_\ell$ for some $\ell$, we must have $\varepsilon_{\ell-1}=\varepsilon_\ell$. For otherwise, a cancellation occurs and contradicts the assumption that $w$ is a reduced word in $\hat{\calG} \cup \hat{\calG}^{-1}$.

Now, regarded as an 