Timeline for For what finite groups is the cardinality of a minimal generating set well defined?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2012 at 22:55 | comment | added | Tsuyoshi Ito | More on “Now, it is known …” part: mathoverflow.net/questions/82603/… | |
| Jun 7, 2011 at 23:00 | comment | added | Frieder Ladisch | I think the argument extends to show that $G$ modulo the commutator subgroup must be a $p$-group. | |
| Jun 4, 2011 at 19:19 | comment | added | Yiftach Barnea | For a nilpotent group $G$ a set $S$ is generating $G$ iff it does modulo the commutator subgroup. Hence, it is reduced to the abelian case. | |
| Jun 4, 2011 at 18:36 | comment | added | Hugo Chapdelaine | I should have thought about it! Thanks! | |
| Jun 4, 2011 at 16:37 | history | answered | user9072 | CC BY-SA 3.0 |