Timeline for Number of countable torsion-free groups
Current License: CC BY-SA 3.0
14 events
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| Nov 12, 2017 at 19:02 | comment | added | YCor | @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 $c=2^{\aleph_0}$. | |
| Nov 12, 2017 at 17:38 | vote | accept | Dominic van der Zypen | ||
| Nov 12, 2017 at 15:49 | review | Close votes | |||
| Nov 13, 2017 at 5:19 | |||||
| Nov 12, 2017 at 15:39 | review | Low quality posts | |||
| Nov 12, 2017 at 15:42 | |||||
| Nov 12, 2017 at 15:34 | history | edited | YCor | CC BY-SA 3.0 |
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| Nov 12, 2017 at 15:33 | comment | added | YCor | 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. | |
| Nov 12, 2017 at 15:26 | comment | added | Péter Komjáth | I think Shelah proved in "Infinite abelian groups, whitehead problem and some constructions" Isr. J. Math., 18(1974), 243–256, that for every infinite cardinal $\kappa$ there are $2^\kappa$ nonisomorphic (and much more) Abelian torsion-free groups of cardinality $\kappa$. The countable case was proved by de Groot in "Indecomposable Abelian groups", Proc. Nederl. Acad. Wet.60(1957), 137-145. | |
| Nov 12, 2017 at 14:27 | answer | added | Jason Starr | timeline score: 10 | |
| Nov 12, 2017 at 13:46 | comment | added | Jason Starr | Here is the fixed link: mathoverflow.net/questions/238664/… | |
| Nov 12, 2017 at 13:46 | comment | added | Jason Starr | @DerekHolt I think that link might be broken. | |
| Nov 12, 2017 at 13:45 | comment | added | Derek Holt | Yes. See YCors's answer to mathoverflow.net/questions/238664 | |
| Nov 12, 2017 at 13:44 | comment | added | Jason Starr | 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. | |
| Nov 12, 2017 at 13:37 | comment | added | Francesco Polizzi | Is not $G$ usually called a torsion-free group? | |
| Nov 12, 2017 at 12:42 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |