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Apr 20, 2023 at 11:29 history edited Sean Eberhard CC BY-SA 4.0
Improve to quadratic in the class-2 case, modulo a guess about the quadratic form xy-zw.
Apr 20, 2023 at 3:03 comment added dennis So beautiful! Thank you for your nice answer!
Apr 20, 2023 at 2:57 vote accept dennis
Apr 19, 2023 at 17:54 comment added Sean Eberhard @SaliniMendisi If $G$ is finitely generated and class-2 nilpotent then $G'$ is also finitely generated. If $g_1, \dots, g_k$ are generators for $G$ then $[g_i, g_j]$ $(1 \le i < j \le k$) are generators for $G'$. If there is no commutator of infinite order then $G'$ is finite, which implies $Z(G)$ has finite index in $G$.
Apr 19, 2023 at 15:54 comment added Salini Mendisi Why does 2-nilpotent and not virtually abelian imply that there is a commutator of infinite order?
Apr 19, 2023 at 11:27 history edited Sean Eberhard CC BY-SA 4.0
add comments on the class-2 case.
Apr 19, 2023 at 10:37 history answered Sean Eberhard CC BY-SA 4.0