Suppose we have a group $G$ acting on a tree $T$ without inversions (i.e. it is never the case that $g \in G$ flips an edge). Let $b, c \in G$ such that $b$ and $c$ individually fix at least 1 vertex, but there is no vertex fixed by both $b$ and $c$. I want to show that $bc$ doesn't fix any vertex of $T$.
I have absolutely no idea how to approach this. I think the easiest way would be to assume there exists $v$ such that $bcv = v$, then somehow make this mean that $b$ and $c$ both fix the same vertex in all cases. If $cv = v$, then $bv = v$, this is a contradiction. Otherwise, $cv \neq v$ and I have no idea where to go from here.
There is a similar setup in another question about groups acting on trees without inversions and vertex stabilisers: A group acting on tree without inversion acts on a line as a translation?, but I don't think the fact proved there is actually useful since we can only conclude from that that such lines $L$ exist or that some element does fix some vertex, we don't have the other direction.
If we had the other direction, then we could show that $bc$ acts on some line $L$ by translations so doesn't fix any vertex. But I'm not sure this converse is true...
Sorry for the jumble of ideas, any help appreciated.