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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 5, 2012 at 18:07 comment added Nik Weaver (In both cases, you can find a counterexample where $S$ contains no comparable elements, hence $S = S^m = S_m$ trivially.)
Aug 5, 2012 at 15:44 comment added Nik Weaver No in both cases. Again, check the commutative case first!
Aug 5, 2012 at 14:36 history answered Masayoshi Kaneda CC BY-SA 3.0