I am trying to understand Example 9.2.5 in Beardon's book The Geometry of Discrete Groups. The goal is to construct a fundamental domain of a Fuchsian group which is not locally finite.
Definitions used by Beardon: A fundamental domain is a connected open subset $D \subseteq H^2$ such that (1) for some $S$ satisfying $D \subseteq S \subseteq \overline{D}$, we have $\bigcup_{g \in G} gS = H^2$ and $S \cap gS = \varnothing$ if $g \ne 1$; and (2) hyperbolic area of $\partial D$ is 0. ($\overline{D}$ means the closure of $D$ in $H^2$.) The fundamental domain $D$ is said to be locally finite if for each compact subset $X \subseteq H^2$, we have $gX \cap \overline{D} \ne \varnothing$ for only finitely many $g \in G$.
Beardon's counterexample is the group $G$ generated by the transformations of the upper half plane $$f(z) = 2z,\,\,\, g(z) = \frac{3z+4}{2z+3}.$$
He first constructs the fundamental domain $D$ (which is locally finite) as the region bounded (in the Euclidean sense) by the semicircles joining -2 to -1, -1 to 1, 1 to 2, and 2 to -2. (Figure 9.2.3 in the book, see below)
The remainder of the construction is illustrated in the next figure below (Figure 9.2.5 in the book) First, we replace the left half of $D$ by its image under $g$. The new region has a boundary segment $[i,2i]$ as well as the semicircles joining 1 to $g(i)$, then $g(i)$ to $g(2i)$, and $g(2i)$ to 2. Then we replace the hyperbolic triangle $(w,1,2w)$ by its image under $f$, which is $(2w,2,4w)$. Everything okay so far. But then Beardon says:
Each Euclidean segment $[\zeta, 2\zeta]$, where $\zeta$ lies on $|z|=1$ and is strictly between $w$ and $i$, is replaced by the equivalent segment $[\zeta', 2\zeta']$.
($\zeta$ and $\zeta'$ are labeled as $z$ and $z'$ in the image below.)
I don't understand how $[\zeta, 2\zeta]$ is equivalent to $[\zeta', 2\zeta']$ under an element of the group $G$, for all $\zeta$ on the arc between $w$ and $i$. The transformation that takes $[\zeta, 2\zeta]$ to $[\zeta', 2\zeta']$ is multiplication by $(\zeta'/\zeta)$, which varies as $\zeta$ varies, so these transformations can't all be in $G$ (since it's a discrete group). Can someone explain what's going on here?
I think it is just very simple Euclidean geometry: if you take any point on the Euclidean segment $[\zeta', 2\zeta']$ then there exists a dilation by some power of 2 (which is an element of your group $G$) which takes this point to the vertical region between 1 and 2 (and above some arcs). For instance, if $\zeta=w$ as in the picture, then it is dilation by 2. Another way to see this is to take the above vertical region and apply to it dilations of the form $z\mapsto 2^{-n}z$, $n$ is a natural number. The images will cover a part of the right half-plane containing the portion of the annulus satisfying $0\le Re(z)\le 1$.