I have been reading Levitt's paper *Automorphisms of Hyperbolic groups and Graphs of Groups*. I am having some trouble trying to fit all the bits together, and would appreciate some help with this last step.

In the paper, Levitt considers *minimal* graphs of groups, and gives results regarding (a specific subgroup of) the outer automorphism group of the fundamental group of an arbitrary graph of groups. A graph of groups is *minimal* if edge groups are proper subgroups of adjacent vertex groups.

Levitt applies this work to the (canonical) JSJ-decomposition of a (one-ended) hyperbolic group. The "canonical" means that we can forget that we are looking at a specific subgroup of the outer automorphism group: his analysis can be applied to the whole outer automorphism group. Which is good, and is the point.

Now, I (believe that I) understand all of the fiddly bits of this paper. However, I have come across a stumbling block when trying to pin it all together. You see, the canonical JSJ-decomposition Levitt is using is due to Bowditch (from Bowditch's paper *Cut points and canonical splittings of hyperbolic groups*). This decomposition is entirely canonical, unlike the one Sela obtains for the torsion-free case (where it is canonical up to certain "moves"). Bowditch's decomposition is canonical because of the addition of "elementary" vertices, that is, vertices whose stabiliser is virtually cyclic. My problem is as follows: I believe that these vertices imply that the graph of groups is not necessarily minimal.

For example, pin together two hyperbolic triangle groups $H$ and $K$ across an infinite cyclic subgroup, so $G=H_{C_1=C_2}K$, and ensure that these subgroups are malnormal infinite cyclic. Then the JSJ-decomposition of this (one-ended hyperbolic group) looks like $H-C-K$, where the edge stabilisers are $C_1$ and $C_2$, and where $C_1=C=C_2$. A contradiction. No?

I am sure I am just missing something silly, something obvious, something I should have seen a long time ago. However, I have no idea what that could possibly be!