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I would like to understand the sets of geodesics in $BS(1, 2)$, which is described in https://arxiv.org/pdf/1908.05321.pdf, Proposition 3 (page 3).

Write $$ G=B S(1, 2)=\left\langle a, t \mid t a t^{-1}=a^2\right\rangle $$ Let $\mathbb{Z}_2=\left\{x \in \mathbb{Q} \mid k^e x \in \mathbb{Z}\right.$ for some $\left.e \in \mathbb{Z}\right\}$ and consider the semidirect product $\mathbb{Z}_2 \rtimes \mathbb{Z}$, where the action of $\mathbb{Z}$ on $\mathbb{Z}_2$ is multiplication by $2$. Then $B S(1, 2) \cong \mathbb{Z}_2 \rtimes \mathbb{Z}$.

Proposition 3. Let $G=B S(1,2)$. The set of words in the following forms comprise a set of unique geodesic representatives for the elements in $\mathbb{Z}_2$.

2a. $\left\{\epsilon, a^{ \pm 1}, a^{ \pm 2}, a^{ \pm 3}\right\}$

2b. $\left\{a^{x_0} t a^{x_1} t \cdots a^{x_d} t^{-d}|d \geq 1,| x_d \mid \in\{2,3\}, A\right\}$

2c. $\left\{t^{-b} a^{x_0} t a^{x_1} \cdots t a^{x_d} t^{-c}\left|b, c, d \geq 1, d=b+c, x_0 \neq 0,\right| x_d \mid \in\{2,3\}, A\right\}$

2d. $\left\{t^{-d} a^{x_0} t \cdots t a^{x_d} \mid d \geq 1, x_0 \neq 0, A\right\}$

Here, $A$ signifies the conditions $\left|x_i\right| \leq 1$ for $i<d$, if $x_{i-1} \neq 0$ then $x_i=0$ for $i<d$, if $x_d>0$ then $x_{d-1} \geq 0$, and if $x_d<0$ then $x_{d-1} \leq 0$.

If I understand the condition $A$ correctly, $w= a^{-1}ta^{0}ta^2 t^{-2}$ is an element in the set described in (2b), but $w$ can also be represented by $ a^1 t a^3 t^{-1}$ which has a shorter length. Did I misunderstand something here?

Thank you for reading. (I have copied this question from StackExchange)

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    $\begingroup$ It's so strange to write $\mathbf{Z}_2$ to mean $\mathbf{Z}[1/2]$. The notation $\mathbf{Z}_2$ is very widely used to mean the $2$-adics, in which all primes are invertible except precisely 2... $\endgroup$
    – YCor
    Commented Oct 6, 2023 at 14:04
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    $\begingroup$ You are right, the second word has a shorter length than the first word and they represent the same group element. Maybe you should contact the authors by email. I consider that a better etiquette than posting here anyway. This question about potential errors in their paper might not be noticed by the authors. $\endgroup$ Commented Oct 6, 2023 at 14:40

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