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Jan 24, 2011 at 21:15 comment added John Klein This gives it for countably generated ones. But a similar argument works I think by filtering $A$ by its poset $\cal P_A$ of finite subgroups and inclusions, and then noting that $\cal P_A$ is contractible (since if $B_\alpha \subset A$ is a finite diagram of finite subgroups, then the subgroup generated by all of these is still finite; this shows that $\cal P_A$ is contractible.
Jan 24, 2011 at 20:29 comment added Jeff Strom Or is this not a problem because I'm working only with abelian groups?
Jan 24, 2011 at 20:25 comment added Jeff Strom Doesn't this just get the result for locally finite groups $A$?
Jan 24, 2011 at 19:50 history answered John Klein CC BY-SA 2.5