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Uncountable groups have many nonconjugate subgroups. (English) Zbl 0629.20001
The whole of this substantial paper is devoted to the proof of the theorem: Main theorem. If G is a group of cardinality $$\lambda$$, $$\lambda$$ an uncountable cardinal, and $$\mu =Min\{\mu:$$ $$2^{\mu}\geq \lambda \}$$, then $$nc_{\leq \mu}(G)\geq \lambda$$. (Where $$nc_{\leq \mu}(G)$$ is the number of pairwise nonconjugate subgroups of G of power $$\mu$$.) This theorem has the interesting conclusion: Conclusion. If $$\lambda$$ is an uncountable cardinal and G a group of cardinality $$\lambda$$ then G has at least $$\lambda$$ pairwise nonconjugate subgroups of power less than $$\lambda$$. This paper completes the work begun by the author in a previous paper [Algebra Univers. 16, 131-146 (1983; Zbl 0521.20015)] where the result was proved under GCH and for many values of $$\lambda$$ (for every G). The techniques used are those of mathematical logic, and the author thoughtfully provides an appendix for nonlogicians containing the facts from mathematical logic that are required for an understanding of the paper.
Reviewer: Sh.Oates-Williams

##### MSC:
 20A15 Applications of logic to group theory 20E07 Subgroup theorems; subgroup growth 03E55 Large cardinals
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##### References:
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