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Jul 6, 2014 at 21:42 comment added YCor OK, but once we are reduced to finitely generated quasi-finite groups, this discussion is of little relevance.
Jul 6, 2014 at 21:37 comment added H. Khas I'm not sure but if there is a sequence of neighborhoods of $1$ which make a base around $1$ in a group topology the topology must be pseudometrizible. If Hausdorff, then metrizable.
Jul 6, 2014 at 21:33 comment added YCor I'm not sure what you mean by "no nondiscrete metric". What do you require about the metric?
Jul 6, 2014 at 21:28 comment added H. Khas I think it admits no nondiscrete Hausdorff topology is equivalent to it admits no nondiscrete metric. It seems such a group cannot have an infinite chain of (normal) subgroups. It suggests to me (somehow) that Hausdorffness may have a better substitution. Btw, I hope I can find a relation to Banach measure.
Jul 6, 2014 at 19:31 history answered YCor CC BY-SA 3.0