Timeline for Nilpotency of a group by looking at orders of elements
Current License: CC BY-SA 2.5
6 events
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| Feb 16, 2011 at 16:25 | comment | added | BS. | @Tom de Metds : yes : if $k,l$ are coprime, the set $P_{kl}$ of elements of order $kl$ is in bijection with the subset of $P_k\times P_l$ consisting of commuting pairs. So the multiplicativity condition says that all such pairs commute, hence the group is the direct product of its Sylows. | |
| Feb 16, 2011 at 9:57 | comment | added | Tom De Medts | @F. Ladish: Thanks -- I wasn't aware of this fact! I even have to confess that I don't see right away why the multiplicativity of $e_G$ implies the nilpotency of $G$. Is there a quick and easy argument? | |
| Feb 13, 2011 at 17:34 | comment | added | Frieder Ladisch | @Tom de Medts: You probably know this, but I mention it nonetheless: Let $e_G(k)$ denote the number of elements of order $k$ in $G$. Since $G$ is nilpotent iff it's the direct product of its Sylow subgroup, $G$ is nilpotent iff $e_G$ is multiplicative on coprime numbers. On the other hand, if you know just the set of occuring orders (without multiplicities), then you can't know: For example, the groups $C_6\times C_2$ and $D_{12}=C_6:C_2$ both have elements of orders 1, 2, 3, 6, only one of them being nilpotent. | |
| Feb 13, 2011 at 8:36 | comment | added | Tom De Medts | That's interesting, thanks! Since you wrote that "nilpotency seems to be the only easy case", do you mean that there exists some other known formula (or algorithm) to determine whether a group is nilpotent, only depending on the orders of the elements? | |
| Feb 11, 2011 at 1:52 | comment | added | Lubin | Welcome to MathOverflow! | |
| Feb 10, 2011 at 23:08 | history | answered | Robert Guralnick | CC BY-SA 2.5 |