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Does the abelian group $\mathbb{Z}[\frac{1}{2}]$ have uncountably many subgroups?

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    @m.k.: It means that we take the subring of the rationals $\mathbb{Q}$ generated by $\mathbb{Z}$ and the element $\frac{1}{2}$.2012-02-22

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No. The strict subgroups are of the form $a\cdot 2^m\mathbb Z \; (m\in \mathbb Z \;,\; a\in 2\mathbb N+1)$.

[Core of proof: look for elements in the subgroup with smallest possible power of two ($=m $) . If there are some take the one with least positive possible odd $a$. Else the subgroup is not strict.]

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    Dear @Jack: you are absolutely right and I have now edited my answer. Thanks$a$lot for spotting and correcting my former erroneous statement.2012-02-22