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17.89 (2010)

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By Bousfield’s theorem, the free pronilpotent completion of a non-cyclic free group has uncountable Schur multiplier. Is it true that the free prosolvable completion (that is, the inverse limit of quotients by the derived subgroups) of a non-cyclic free group has uncountable Schur multiplier?

Contributor: R. Mikhailov

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