Prove that certain group has exponential growth

Question

Let $R$ be the ring of dyadic rationals. I'm trying to prove that $$G = \left\{\begin{bmatrix} 2^n & 0 \\ r & 1 \end{bmatrix}, r \in R, n \in \mathbb{Z}\right\}$$ has exponential growth. I've shown that $G$ is generated by $s := \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, t:= \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ and that $t^{-1}st=s^2$. I've also shown, that $G=K\rtimes D$(semidirect product) where $D$ are the diagonal Matrices in $G$ and $K$ is the kernel of the HM $\begin{bmatrix} 2^n & 0 \\ r & 1 \end{bmatrix} \mapsto n$. $K$ is obviously isomorphic to $R$ so not finitely generated. I'd be really grateful for a nudge in the right direction, why these facts would imply $G$ to be of exponential growth.

Answer 1

The quickest way I know to do this example is to show that there is a linearly growing function $L(n)$ such that for each $i=1,\ldots,2^n$, the group element $s^n = \begin{bmatrix} 1 & 0 \\ 2^n & 1 \end{bmatrix} $ has word length $\le L(n)$.

Given any exponent $i=1,...,2^n$, letting $k = \lfloor j/2 \rfloor$, if $i=2k$ is even then $t^{-1} s^k t = s^j$, whereas if $i=2k+1$ is odd then $t^{-1} s^k t s = s^j$. In either case the word lengths of $s^{j}$ and of $s^k$ differ by at most $3$, and $0 \le k \le 2^{n-1}$. If $k=0$ one is done, and if $1 \le k \le 2^{n-1}$ then one continues inductively.

So one can use $L(n)=3n$ for the desired bound.