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Does there exist an uncountable group , every proper subgroup of which is countable ?

user26857
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  • I think I have a vague idea about how Shelah constructed his Jonsson group on $\aleph_1$. But (1) it is quite a complicated construction (as fitting a theorem of Shelah); and (2) it's only a vague idea, and I've made enough mistakes for one day. So I'll just let Derek Holt post an answer with the reference that he found, and remark that without the axiom of choice it is indeed consistent. – Asaf Karagila Apr 09 '15 at 12:24
  • By using inverse limit of groups, such groups can be created. When I remember, I will write such one. – mesel Apr 09 '15 at 12:32

1 Answers1

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The existence of such groups is apparently proved in the paper Shelah, Saharon "On a problem of Kurosh, Jónsson groups, and applications". Word problems, II (Conf. on Decision Problems in Algebra, Oxford, 1976), pp. 373–394, Stud. Logic Foundations Math., 95, North-Holland, Amsterdam-New York, 1980.

I haven't seen the paper itself, but here is the description in Math. Reviews from MathSciNet.

Using small cancellation theory, the author constructs some remarkable infinite groups, notably "Jónsson groups'', that is, groups of uncountable cardinality $\lambda$ containing no proper subgroups (or better: no proper subsemigroup) of the same cardinality. One construction works for all successor cardinals if we assume the generalized continuum hypothesis; a variant works for $\aleph_1$ with no set-theoretic assumptions.

The groups constructed actually have slightly stronger properties and hence constitute counterexamples to a few not obviously related conjectures, notably (Theorem C): The continuum hypothesis implies that there is an uncountable group which admits no nontrivial topology (i.e. cannot be made into a topological group in a nontrivial way).

Derek Holt
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    Scott, W. R. "Groups and cardinal numbers." Amer. J. Math. 74, (1952). 187-197. (Also http://www.ams.org/mathscinet-getitem?mr=46364) proved that for abelian groups, this is false. Lauchli proved, and I refurbished that proof and extended it a bit in my masters, that without the axiom of choice it is possible that an abelian group is uncountable but only have countable proper subgroups. – Asaf Karagila Apr 09 '15 at 13:00
  • Google found a textbook presentation in Geometry of Defining Relations in Groups, by Ol'shanskii (around p. 391) – the gods from engineering Apr 09 '15 at 14:38