2008
DOI: 10.1515/jgt.2008.025
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Reflection principle characterizing groups in which unconditionally closed sets are algebraic

Abstract: We give a necessary and sufficient condition, in terms of a certain reflection principle, for every unconditionally closed subset of a group G to be algebraic. As a corollary, we prove that this is always the case when G is a direct product of an Abelian group with a direct product (sometimes also called a direct sum) of a family of countable groups. This is the widest class of groups known to date where the answer to the 63 years old problem of Markov turns out to be positive. We also prove that whether every… Show more

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Cited by 21 publications
(33 citation statements)
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“…Our next proposition provides an easy algorithm for verifying whether an abelian group is an M -group. Its proof, albeit very short, is based on the fundamental fact that all unconditionally closed subsets of abelian groups are algebraic [10,Corollary 5.7]. Proof.…”
Section: Markov's Problem For Abelian Groupsmentioning
confidence: 99%
See 1 more Smart Citation
“…Our next proposition provides an easy algorithm for verifying whether an abelian group is an M -group. Its proof, albeit very short, is based on the fundamental fact that all unconditionally closed subsets of abelian groups are algebraic [10,Corollary 5.7]. Proof.…”
Section: Markov's Problem For Abelian Groupsmentioning
confidence: 99%
“…Proof. According to [10,Corollary 5.7] and [11, According to the Prüfer theorem [15,Theorem 17.2], a non-trivial abelian group G of finite exponent is a direct sum of cyclic subgroups…”
Section: Markov's Problem For Abelian Groupsmentioning
confidence: 99%
“…The family of unconditionally closed subsets of G coincides with the family of closed subsets in another T 1 topology on G, namely, the infimum of all Hausdorff group topologies on G. This topology was explicitly introduced in [7] under the name Markov topology.…”
Section: Problem 12 Find Necessary and Sufficient Conditions For Thmentioning
confidence: 99%
“…In 1944 Markov [76] asked if the equality Z G = M G holds for every group G. He himself showed that the answer is positive in case G is countable [76]. Moreover, in the same manuscript Markov attributes to Perel'man the fact that Z G = M G for every Abelian group G (a proof has never appeared in print until [37]). An example of a group G with Z G = M G was given by Gerchard Hesse [66].…”
Section: The Zariski Topology and The Markov Topologymentioning
confidence: 99%