Group acting on trees with relations $abc=1$ and $cba=1$

Question

Suppose $G$ is a finitely presented group that acts on a tree $T$ by isometries, and let $a,b,c\in G$ with relations $abc=1$ and $cba=1$. If two of $a,b,c$ are hyperbolic, does this imply the third element is also hyperbolic? Or any additional conditions that we need to make the implication hold?

Answer 1

Assuming $a$ is hyperbolic, and letting $b=a^{-1}$, it follows that $b$ is hyperbolic too. The relation $ab=c^{-1}$ tells you that $c=1$ which is not hyperbolic.

To say something more general, let $a$ be hyperbolic with axis $A_a \subset T$, and $b$ be hyperbolic with axis $A_b \subset T$.

Suppose that $\alpha = A_a \cap A_b$ is a segment of positive length $L = \text{length}(\alpha)$ (which might be infinite). If the translation lengths of $a$ and $b$ are equal to each other and both less than $L$, and if $a$ and $b$ move points on $\alpha$ in opposite directions, then $ab=c^{-1}$ will fix a point on $\alpha$ and therefore $c$ will not be hyperbolic.

On the other hand suppose $A_a \cap A_b$ is either a single point or the empty set. In this case $ab=c^{-1}$ is hyperbolic and therefore $c$ is hyperbolic.

If you want to read more about this stuff I suggest the paper of Culler and Morgan, "Group actions on $\mathbb R$-trees".