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Oct 8, 2018 at 9:48 comment added Andy Ah ok, just figured this out. If $g\in G$ is not in the center, there is some $h\in G$ with which it does not commute. Then $C_h$ is a finite index normal subgroup avoiding $g$.
Oct 8, 2018 at 9:42 comment added Andy Why is $G/Z(G)$ residually finite when $G$ is FC?
Aug 7, 2014 at 12:38 vote accept Robin Tucker-Drob
Aug 7, 2014 at 11:40 answer added YCor timeline score: 7
Aug 7, 2014 at 9:52 comment added YCor Obviously $G/Z(G)$ is residually finite, hence MAP, so it works when $Z(G)$ is trivial. In general it sounds tricky.
Aug 7, 2014 at 3:55 history asked Robin Tucker-Drob CC BY-SA 3.0