Is it true that there exist $2^{\aleph_0}$ pairwise non-isomorphic torsion-free countable groups?
$\begingroup$
$\endgroup$
8
asked Nov 12, 2017 at 12:42
-
5$\begingroup$ Is not $G$ usually called a torsion-free group? $\endgroup$– Francesco PolizziCommented Nov 12, 2017 at 13:37
-
4$\begingroup$ For every subset $S$ of the set $P$ of positive, prime integers, let $G_S\subset \mathbb{Q}$ denote the subgroup of those fractions whose denominator is divisible by $p$ only if $p$ is in $S$. The subset $S$ is uniquely recovered from $G_S$ as the set of primes $p$ such that the "multiplication by $p$" map on $G_S$ is an isomorphism. $\endgroup$– Jason StarrCommented Nov 12, 2017 at 13:44
-
5$\begingroup$ Yes. See YCors's answer to mathoverflow.net/questions/238664 $\endgroup$– Derek HoltCommented Nov 12, 2017 at 13:45
-
3$\begingroup$ Exercise: (a) show that $\mathrm{Aut}(\mathbf{Q}^2)$ is countable. (b) any isomorphism between two subgroups of $\mathbf{Q}^2$ extends to an automorphism of $\mathbf{Q}^2$. (c) show that for any prime $p$, the number of subgroups of $\mathbf{Z}[1/p]^2$ is $2^{\aleph_0}$ (d) Conclude that $\mathbf{Z}[1/p]^2$ has $2^{\aleph_0}$ non-isomorphic subgroups. $\endgroup$– YCorCommented Nov 12, 2017 at 15:33
-
1$\begingroup$ @PéterKomjáth: the classification of subgroups of $\mathbf{Q}$ can be found in Ross A. Beaumont and H. S. Zuckerman, A characterization of the subgroups of the additive rationals, Pacific J. Math. Volume 1, Number 2 (1951), 169-177. That there are $2^{\aleph_0}$ non-isomorphic such groups follows (and is immediate anyway). The 1957 paper by de Groot is something more difficult with a construction of $2^c=2^{2^{\aleph_0}}$ non-isomorphic torsion-free abelian groups of cardinal $c=2^{\aleph_0}$. $\endgroup$– YCorCommented Nov 12, 2017 at 19:02
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
1
Extended cw answer based on Jason Starr's comment.
The additive group of rationals admits $2^{\aleph_0}$ non-isomorphic subgroups.
Denote by $P\subset \mathbb{Z}_{>0}$ the set of positive, integer primes. This is a countably infinite set by Euclid's proof of the infinitude of primes. The set $\mathcal{G}$ of saturated, multiplicatively closed subsets $S$ of $\mathbb{Z}$ is in bijection with the power set $\mathcal{P}(P)$ by the rule $S\mapsto S\cap P.$
For every saturated, multiplicatively closed subset $S$ of $\mathbb{Z},$ denote by $G_S$ the fraction ring, $$G_S =S^{-1}\mathbb{Z} \subset \mathbb{Q}.$$ This is a subring of the countably infinite ring $\mathbb{Q}$, thus also $G_S$ is countably infinite. Moreover, the subset $$ \{p\in P |\ \forall x\in G_S, \ \exists y\in G_S, \ p\cdot y= x\}$$ equals $S\cap P.$ Thus, if $G_S$ is isomorphic to $G_T$ as Abelian groups, then $S$ equals $T$. Therefore, the collection of Abelian groups $G_S$ is a system of pairwise non-isomorphic, countably infinite, torsion-free groups that are indexed by the set $\mathcal{G}$ with cardinality $2^{\aleph_0}.$
-
2$\begingroup$ This can also be stated as: the ring $\mathbf{Q}$ possesses $2^{\aleph_0}$ subrings (these are precisely the $\mathbf{Z}[S^{-1}]$ when $S$ ranges over subset of the set of primes) and these are pairwise non-isomorphic as additive groups (because for every prime $p$, we have $p\in S$ iff multiplication by $p$ is surjective in $\mathbf{Z}[S^{-1}]$). $\endgroup$– YCorCommented Nov 12, 2017 at 16:14