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Sep 20, 2011 at 19:34 comment added George Lowther @ARupinski: $G$ is not a countable union of subgroups isomorphic to images of $S_\infty$.
Sep 20, 2011 at 16:12 vote accept ARupinski
Sep 20, 2011 at 16:12 comment added ARupinski Your point 5 seems to confirm what I suspected but could not rigorously prove: that there must be more to $G$ than just a bunch of copies of $S_\infty$.
Sep 20, 2011 at 7:47 history answered Amit Kumar Gupta CC BY-SA 3.0