finite nilpotent groups
Finite (http://planetmath.org/Finite) -groups are nilpotent.
Example. Infinite -groups may not always be nilpotent. In the extreme there are counterexamples like the Tarski monsters . These are infinite -groups in which every proper subgroup has order . Therefore given any two non-trivial elements in which generate . In particular, the only central element is 1 so that the upper central series is trivial and therefore is not nilpotent.
Indeed, Tarski monsters are not in fact solvable groups which is a weaker property than nilpotent.
Example. Some infinite -groups are nilpotent. Indeed, some infinite -groups are even abelian such as – the countable dimension vector space over the field – and the Prüfer group – the inductive limit of .
If is a finite group and a Sylow -subgroup of then
Now we prove Theorem 2
(1) implies (2). Suppose that is nilpotent and that is a Sylow -subgroup of . Then as is nilpotent, every subgroup of is subnormal in , meaning, if is properly contained in then properly contains . Thus is larger than or . However because is a Sylow -subgroup we know so we conclude . Therefore every Sylow -subgroup of is normal in .
(2) implies (3). Suppose every Sylow subgroup of is normal in . Then by the Sylow theorems we know that for every prime dividing there is exactly one Sylow -subgroup of – as all Sylow -subgroups are conjugate and here by assumption all are also normal.
(3) implies (4). Suppose that there is a unique Sylow -subgroup of for every . Then by the Sylow theorems every Sylow subgroup of is normal in . Furthermore, if and are two distinct Sylow subgroups then they they are Sylow subgroups for different primes so that by Lagrange’s theorem their intersection is trivial. Let the Sylow subgroups of . Then as each is normal in we have and we have also demonstrated for therefore is the direct product of .
|Title||finite nilpotent groups|
|Date of creation||2013-03-22 15:46:40|
|Last modified on||2013-03-22 15:46:40|
|Last modified by||Algeboy (12884)|