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13.46 (1995)

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Can every uncountable abelian group of finite odd exponent be partitioned into two subsets so that neither of them contains cosets of infinite subgroups? Among countable abelian groups, such partitions exist for groups with finitely many involutions.

Contributor: I. V. Protasov

Progress

Yes, it can (E. G. Zelenyuk, Math. Notes, 67 (2000), 599–602).

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