I'm reading the book "The Banach-Tarski Paradox" by Stan Wagon. The Theorem 1.5 (page no:7) it states the following:

Theorem 1.5 (AC). $S^1$ is countably $SO_2$-paradoxical. If $G$ denotes the group of translations modulo $1$ acting on $[0,1)$, then $[0,1)$ is countably $G$-paradoxical.

But I can't understand what is meant here by **"$G$ denotes the group of translations modulo $1$". What group is this?

Please give its definition. Thank you.

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    $\begingroup$ Consider the folliwing operation on $I = [0, 1)$. If $x , y \in I$, then $x \oplus y = x + y$ when $x + y \in I$, otherwise $x \oplus y = x + y - 1$. Then the group of translations modulo $1$ is given by the transformations on $I$ of the form $t(y) : x \mapsto x \oplus y$, for $y \in I$. $\endgroup$ – Andreas Caranti Apr 2 '18 at 9:44
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    $\begingroup$ This group is isomorphic to $S^{1}$ (regarded as a subgroup of the multiplicative group of non-zero complex numbers) under the isomorphism sending $t(y)$ to $e^{i 2 \pi y}$. $\endgroup$ – Andreas Caranti Apr 2 '18 at 9:47
  • $\begingroup$ @ A Caranti...Oh..thanks a lot. The group of translation mod 1, $G=\{t_y|y \in [0,1)\}$ is isomorphic to $S^1$ as you mentioned. $\endgroup$ – Indrajit Ghosh Apr 2 '18 at 13:15

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