finite nilpotent groups
The study of finite nilpotent groups mostly centers around the study of -groups. This is because of the following two theorems.
Theorem 1.
Finite (http://planetmath.org/Finite) -groups are nilpotent.
Proof.
From the class equation we know the center of a finite -group is non-trivial. Thus by induction the upper central series of a -group terminates at . So is 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 .
Theorem 2.
Let be a finite group. Then all the following are equivalent.
-
1.
is nilpotent.
-
2.
Every Sylow subgroup of is normal.
-
3.
For every prime , there exists a unique Sylow -subgroup of .
-
4.
is the direct product of its Sylow subgroups.
For the proof recal the following consequence of the Sylow theorems:
Proposition 3.
If is a finite group and a Sylow -subgroup of then
(See Subgroups Containing The Normalizers Of Sylow Subgroups Normalize Themselves)
Now we prove Theorem 2
Proof.
(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 .
(4) implies (1). Suppose that is a product of its Sylow subgroups. Then as every Sylow subgroup is a -group, is a product of nilpotent groups so itself is nilpotent. ∎
Title | finite nilpotent groups |
---|---|
Canonical name | FiniteNilpotentGroups |
Date of creation | 2013-03-22 15:46:40 |
Last modified on | 2013-03-22 15:46:40 |
Owner | Algeboy (12884) |
Last modified by | Algeboy (12884) |
Numerical id | 18 |
Author | Algeboy (12884) |
Entry type | Topic |
Classification | msc 20D15 |
Classification | msc 20E15 |
Classification | msc 20E34 |
Related topic | NilpotentGroup |
Related topic | DirectProductAndRestrictedDirectProductOfGroups |
Related topic | ClassificationOfFiniteNilpotentGroups |