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Jul 20, 2020 at 18:17 vote accept R. Srivastava
Jul 20, 2020 at 14:40 history closed YCor
LeechLattice
HenrikRüping
M. Winter
Derek Holt
Not suitable for this site
S Jul 20, 2020 at 14:30 history suggested Carl-Fredrik Nyberg Brodda
Not part of algebraic geometry.
Jul 20, 2020 at 13:45 review Suggested edits
S Jul 20, 2020 at 14:30
Jul 20, 2020 at 12:21 answer added Claus timeline score: 5
Jul 20, 2020 at 12:19 history edited R. Srivastava CC BY-SA 4.0
deleted 16 characters in body; edited title
Jul 20, 2020 at 8:30 history edited R. Srivastava CC BY-SA 4.0
added 139 characters in body
Jul 20, 2020 at 8:19 comment added Claus @R. Srivastava Rajeev welcome to MO. Maybe best if you specify where you have read this. This might make it easier to give you an answer
Jul 20, 2020 at 8:07 comment added HenrikRüping I would read the informal sentence "cannot be distinguished by only using the group structure" as that there is a group isomorphism $G\rightarrow G$ sending one element to the other. That isomorphism sending $x$ to $gxg^{-1}$ is given by conjugation with $g$. Certainly a set invariant under conjugation spans a normal subgroup. Certainly conjugacy classes are not subgroups themselves, they don't need to contain the neutral element.
Jul 20, 2020 at 7:32 review Close votes
Jul 20, 2020 at 14:48
Jul 20, 2020 at 7:19 review First posts
Jul 20, 2020 at 7:38
Jul 20, 2020 at 7:13 history asked R. Srivastava CC BY-SA 4.0