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Apr 30, 2021 at 15:12 vote accept CommunityBot
Apr 22, 2021 at 2:25 history became hot network question
Apr 21, 2021 at 21:27 comment added Asaf Karagila There are explicit polynomials in $\Bbb Z[x_0,\dots]$ which have roots if and only if ZFC is inconsistent. Surely you can use one of them to engineer a group that is trivial if and only if ZFC is inconsistent.
Apr 21, 2021 at 19:21 answer added Joel David Hamkins timeline score: 21
Apr 21, 2021 at 19:15 answer added IJL timeline score: 2
Apr 21, 2021 at 18:37 comment added user178109 @LSpice by explicit I mean state the number of generators and then write down the relations
Apr 21, 2021 at 18:30 comment added LSpice It depends what 'explicit' means; you can define $G = \langle a\rangle/N$, where $N = \langle a\rangle$ or $N = 1$ depending on whether the continuum hypothesis is true.
Apr 21, 2021 at 18:25 history asked user178109 CC BY-SA 4.0