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Aug 20, 2019 at 21:59 comment added Derek Holt @YCor Oh dear yes, I am getting tired!
Aug 20, 2019 at 21:30 comment added YCor @DerekHolt OK great. PS François's comment is a joke.
Aug 20, 2019 at 21:15 comment added Derek Holt @YCor Among the groups $G,G^{(2)},\ldots,G^{(k-1)}$, only $G^{(k-1)}$ has a direct summand of order $2$.
Aug 20, 2019 at 20:00 comment added YCor Considering $G=\bigoplus_{n\ge 1}G_{kn}$, one gets $G$ isomorphic to $G^{(k)}$; can one ensure periodicity exactly $k$?
Aug 20, 2019 at 19:27 comment added Francois Ziegler @SantanaAfton But you asked for not perfect?!
Aug 20, 2019 at 19:25 comment added Santana Afton This is perfect, thank you!
Aug 20, 2019 at 19:24 vote accept Santana Afton
Aug 20, 2019 at 19:23 history answered Derek Holt CC BY-SA 4.0