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Aug 4, 2018 at 5:19 comment added Abbas Jafarzadeh Yes, Of course, $G$ is an arbitrary group.
Aug 4, 2018 at 5:12 vote accept Abbas Jafarzadeh
Aug 2, 2018 at 11:33 answer added Denis T timeline score: 3
Aug 2, 2018 at 10:13 comment added Derek Holt For finite groups it is not hard to show that both conditions are equivalent to $G$ being nilpotent, but I expect you want the results for all groups.
Aug 2, 2018 at 9:40 history edited Martin Sleziak
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
Aug 2, 2018 at 7:30 review First posts
Aug 2, 2018 at 7:44
Aug 2, 2018 at 7:28 history asked Abbas Jafarzadeh CC BY-SA 4.0