# Cardinality of the set of countable dense subgroups of the reals up to isomorphism.

Joel David Hamkins in an answer to my question Countable Dense Sub-Groups of the Reals points out that "one can find an uncountable chain of countable dense additive subgroups of $\mathbb{R}$ whose subset relation has the order type of the continuum $\langle \mathbb{R},<\rangle$."

I would like to know what is the cardinality of the set of countable dense additive subgroups of the reals (up to isomorphism)? Is it undecidable?

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The result of Joel quoted above certainly shows a complicated structure, but does not actually provide continuum many non-isomorphic subgroups since there is no reason why $G$ being a subgroup of $H$ should imply that $G$ and $H$ are not isomorphic. Indeed, considering two groups generated by the rationals and two infinite algebraically independent sets of reals will provide counterexamples to this.

However, the structure theorem of rank one torsion free abelian groups does shows there are continuum many such. For example, for any set of primes $P$ let $G_P$ be the set of all rationals whose denominator is a product of primes only from $P$. It is easy to see that every element has $p^\text{th}$ roots if $p \in P$ but that $1$ has no $q^\text{th}$ root if $q\notin P$. Hence all these groups are non-isomorphic.

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A countable abelian group can be encoded by a presentation. Such a presentation has countably many generators, and countably many relations (each one being a finite word in the generators).

The isomorphism type of the group is encoded by a countably infinite word in a countable alphabet. The cardinality of those is equal to the continuum. Hence, there are at most continuum many isomorhpism types. Joel David Hamkins has already provided a lower bound, therefore, there are exactly continuum many isomorhpism types.

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There are continuum many functions ${\cdot}\colon\omega\times\omega\to\omega$, and therefore at most continuum groups with a fixed countable domain. No need to involve presentations in this argument. –  Emil Jeřábek Jun 16 '11 at 13:49
For that matter, the real line has only continuum many countable subsets altogether. –  Andreas Blass Jun 16 '11 at 13:52