Bass-Serre tree of Isom($\mathbb{Z}$) I wish to draw the Bass-Serre tree associated to Isom($\mathbb{Z}$). Now I'll be honest, I'm not absolutely certain what a Bass-Serre tree is, but this is in the context of amalgamated free products, so I'll work from the wiki definition here. So how do we work through this? Do we look at both reflection and translation as isometries on $\mathbb{Z}$ and use this?
 A: The Bass-Serre tree of a free product $G=A\ast B$ is a tree equipped with a specific action of $G$. There are two types of vertices - red vertices and blue vertices. No two vertices of the same colour are incident.
Red vertices correspond to the group $A$, and incident edges are labelled by elements of $A$. The group $A$ acts on the tree by fixing some red vertex $v_A$ and permuting the incident edges (and the corresponding rooted trees) using the right-action (say) of $A$ on the labelling. Every red vertex is fixed by some conjugate of $A$, and every conjugate fixes some red vertex.
Identically, blue vertices correspond to the group $B$, and incident edges are labelled by elements of $B$. The group $B$ acts on the tree by fixing some blue vertex $v_B$ and permuting the incident edges (and the corresponding rooted trees) using the right-action (say) of $B$ on the labelling. 
Every blue vertex is fixed by some conjugate of $B$, and every conjugate fixes some red vertex.
Note that edges are labelled by both $A$ and $B$ - a labelling for each end of the edge. If this was a free product with amalgamation, the action of $A$ and $B$ on the tree would fix certain edges.
Note that the quotient of this tree by the action of your group $G$ is a graph of groups. This is why we say that "$G$ is the fundamental group of a graph of groups" (the Bass-Serre acts like the universal cover.)
In your example, as you showed in the comments, both $A$ and $B$ are cyclic of order two. Therefore, each vertex has precisely two incident edges. Which trees do you know where every vertex has degree precisely two?!?
