My question is the following:

In Bowditch's JSJ-decomposition of hyperbolic groups, can elementary (virtually-cyclic) vertices have degree 1? If not, why not?

I had thought for a long time that this was not possible, as it seems to be pointless (with respect to Outer automorphism groups). For example, if you have a hyperbolic group $A\ast_CB$ where $A$ is virtually cyclic then sinking the vertex corresponding to $A$ into $B$ does not change the information we gain about the outer automorphism group.

Have I been thinking about this incorrectly? Do rigid vertices actually have all their possible virtually cyclic splittings protruding from them like this (as these splittings are not allowed so we have to push them out)? So rigid vertices are surrounded by elementary vertices of degree one which are possible attaching points for other hyperbolic groups? Or is there some subtlety of rigid hyperbolic groups which I have not understood?

EDIT: As the commenter has pointed out, the source of my confusion is the difference between Sela's and Bowditch's definitions. Sela only allows edge groups to be maximal, so I presumed that when you add in the elementary vertices this maximality is somehow maintained. But, seemingly, this is not the case (edge groups have to be maximal in the adjacent non-elementary vertex groups, but not in the elementary ones). Now, I have a single silly question before I think I understand this all better - can I simply attach roots? More precisely, say I have a hyperbolic group and every relator which contains the word $w$ contains it as part of $w^{in}$ for some fixed $n$, so $G=\langle X, w; R_1(X, w^n), \ldots, R_k(X, w^n)\rangle$. Then the word $w$ can be taken out as a root, and we obtain the group $\langle X, u, w; R_1(X, u), \ldots, R_k(X, u), u=w^n\rangle=G_1\ast_{u=w^n}\langle w\rangle$. Does Bowditch's definition encode this?