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Is it true or not $\mathbb Z + \alpha\mathbb Z$ is not isomorphic to $\mathbb Z + \beta\mathbb Z$ as an ordered (totally ordered) subgroup of the real numbers? where $\alpha, \beta$ are different non-zero rational numbers. Thanks

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By the extended Euclidean algorithm, if $n/d$ is written in lowest terms, we can find

$$ u\frac{n}{d} + v = \frac{1}{d}$$

and so

$$\mathbb{Z} + \frac{n}{d} \mathbb{Z} = \frac{1}{d} \mathbb{Z}$$

and multiplication by $d$ gives an isomorphism of ordered groups

$$ \frac{1}{d} \mathbb{Z} \mapsto \mathbb{Z} $$

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    The above calculation is related to the topic of "invertible ideals" which you may or may not find interesting.2012-03-30
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    Thanks Hurkyl for nice argument. Still, I am not clear for ordered abelian group.2012-03-30