The tree of $\langle a,b\mid a^2=b^3\rangle$. 
I want to draw the usual tree that $G=\langle a,b\mid a^2=b^3\rangle$ acts on.

EDIT THIS IS WRONG
So I wrote $G=\mathbb{Z}_2*\mathbb{Z}_3=\langle a\rangle*\langle b\rangle$. I set $G_0=\{e\}$ the trivial group and then, for the tree, $V(x)= \sqcup\{gG_i,\ g\in G, G_i\in\{G_0,\mathbb{Z}_2,\mathbb{Z}_3\}\}$ and $E_+(X)=G\times I$. I also know that

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*Every vertice of the form $xG_0$ has a neighbor of the form $gG_i$ iff $g^{-1}x\in G_i$

*The vertice $xG_0$ has only one neighbor of the form $yG_i$
EDIT The questions remain thw same
1)Now I can't really manage to draw this tree. Any help?


*Is there any book or pdf that contains examples of trees of free products, amalgams and HNN extensions?

Thanks in advance!
 A: Besides Serre's excellent book "Trees", which emphasizes the algebraic point of view, one can read the excellent article "Topological methods in group theory" by Scott and Wall. My answer will present that topological point of view.
The group in question is, as said in the comments, the amalgamated free product $\mathbb Z *_{\mathbb Z} \mathbb Z$ with amalgamating homomorphisms $f_0,f_1 : \mathbb Z \to \mathbb Z$ defined by $f_0(x)=2x$ and $f_1(x)=3x$. First I'll build a topological model for this group. It will be a "graph of spaces" (in the language of Scott and Wall). It has two "vertex spaces, each of which is a circle, call these two circles $C_0,C_1$. It also has an "edge space", which is an annulus $S^1 \times [0,1]$. These are bonded together with two gluing maps, $f_i : S^1 \times \{i\} \to C_i$ for $i=0,1$, where $f_2$ is a 2-fold covering map and $f_3$ is a 3-fold covering map. Altogether, one obtains a 2-dimensional space $X$ which is the quotient space obtained from the disjoint union $C_0 \coprod S^1 \times [0,1] \coprod C_1$, by making the identifications $(x,0) \sim f_0(x)$ and $(x,1) \sim f_1(x)$.
In this space $X$ we define a decomposition into circles: the image of $C_0$ under the quotient is one circle, equal to the image of $S^1 \times \{0\}$; the image under $C_1$ is another circle, equal to the image of $S^1 \times \{1\}$; and for each $t \in (0,1)$ the image of $S^1 \times \{t\}$ is another circle.
Now consider the universal covering space $\widetilde X$, and the covering map $p : \widetilde X \to X$. The decomposition of $X$ itself into circles lifts to a decomposition of $\widetilde X$ into subsets as follows: for each circle $C \subset X$, the subset $p^{-1}(C) \subset \widetilde X$ is a union of path components. The decomposition of $\widetilde X$ is defined to be the collection of path components of $p^{-1}(C)$ as $C$ varies over the decomposition elements of $X$. Let $T$ be the quotient space this decomposition, and let $q : \widetilde X \to T$ be the quotient map.
What Scott and Wall prove is that $T$ is a tree, and this is the tree that you want. There's quite a bit of work left to prove that $T$ is a tree, but here is a detailed intuitive description.
In $\widetilde X$, pick a component of $p^{-1}(C_0)$. That component is a line I'll denote $L_0$ (this is part of what Scott and Wall prove). Projecting $L_0$ to $T$ we get a vertex $v_0$ of $T$.
In $\widetilde X$, there are two "strips" incident to the line $L_0$ which I'll denote $S_{0,1}$ and $S_{0,1}$, each homeomorphic to $\mathbb R \times [0,1]$, each having one boundary component on the line $L_0$, with opposite boundary components being lines denoted $L_{0,1},L_{0,2}$.  Projecting to $T$ we get two edges I'll denote $E_{0,1},E_{0,2}$ incident to $v_0$, with opposite endpoints $v_{0,1}$ and $v_{0,2}$. Thus, $v_0$ is a valence 2 vertex of $T$. 
Incident to the $L_{0,i}$ for each $i=1,2$ there are three "strips", one of which has already been described, namely $E_{0,i}$; let me denote the other two $S_{0,i,1}$ and $S_{0,i,2}$, each with opposite boundary line denoted $L_{0,i,1}$ and $L_{0,i,2}$. Projecting to $T$, incident to each $v_{0,i}$ there are two edges $E_{0,i,1}$, $E_{0,i,2}$ each with opposite vertex denoted $V_{0,i,1}$, $V_{0,i,2}$. 
Now let me just say how to continue, simplifying the picture by describing only $T$, and summarizing the whole construction of $T$ as a definition-by-induction. 


*

*Step 0: Start with the vertex $v_0$. 

*Step 1: Attach two edges, making $v_0$ a valence 2 vertex, and introducing 2 new vertices each having valence 1 so far. 

*Step 2: To each of the 2 new vertices, attach 2 edges, making each of those into valence 3 vertices, and introducing $2*2=4$ new vertices each having valence 1 so far.

*Step 3: To each of the 4 new vertices, attach 1 edge, making each of those into valence 2 vertices, and introducing $2*2*1=4$ new vertices each having valence 1 so far.

*Step 4: To each of the 4 new vertices, attach 2 edges, making each of those into valence 3 vertices, and introducing $2*2*2=8$ new vertices, each having valence 1 so far.


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*Step 2n+1: To each of the $2^{n+1}$ new vertices, attach 1 edge, making each of those into valence 2 vertices, and introducing $2^{n+1}$ new vertices each having valence 1 so far.

*Step 2n+2: To each of the $2^{n+1}$ new vertices, attach 2 edge2, making each of those into valence 3 vertices, and introducing $2^{n+2}$ new vertices each having valence 1 so far.


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When you're done with this inductive definition, you have your tree $T$.
