# Decomposition of a group whose Cayley graph is a tree

This is an exercise taken from Chapter 9 of a French book, Géométrie et Théorie des Groupes. It says, roughly, the following:

Show that a finitely generated hyperbolic group, whose Cayley graph is a tree, is isomorphic to an amalgamated sum of certain number of copies of $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$.

Have I missed something here, but isn't the Cayley graph of $G=$PSL$_2(\mathbb{Z})$ a tree, but $G$ is a free product of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$? Is it possible to rewrite this free product in another way so that the statement of the exercise holds? All I know is that it contains a subgroup isomorphic to the free group of rank two.

• I am probably missing something silly here, but how can a group with an element of order $3$ have a tree for its Cayley graph? Surely the element of order $3$ gives a loop of order $3$ in the graph? – user1729 Mar 31 '15 at 8:40
• @user1729: presumably you don't use the element of order $3$ as one of the generators. – Qiaochu Yuan Mar 31 '15 at 20:57
• @Qiaochu yes, true, but it still corresponds to a loop. If $w$ is an element of order three then it gives a loop in the Cayley graph. (As does any relator). Or, to put it another way, the presentation complex contains $2$-cells so the universal cover (the Cayley complex) cannot have a tree as its 1-skeleton. However, its 1-skeleton is the Cayley graph. – user1729 Apr 1 '15 at 7:07
• Perhaps the OP meant quasi-isometric to a tree? – user1729 Apr 1 '15 at 7:09
• Oh, I see what you mean. Whatever generators you use, there's some word in them which has order $3$... – Qiaochu Yuan Apr 1 '15 at 7:20

$\mathbb{Z}_2 \ast \mathbb{Z}_3$ can be written as an amalgamated sum $\mathbb{Z}_2 \ast_{\mathbb{Z}} \mathbb{Z}$, where the map $\mathbb{Z} \to \mathbb{Z}_2$ is zero and the map $\mathbb{Z} \to \mathbb{Z}$ is multiplication by $3$.