finite nilpotent groups

The study of finite nilpotent groups mostly centers around the study of p-groups. This is because of the following two theorems.

Theorem 1.

Finite ( p-groups are nilpotentPlanetmathPlanetmathPlanetmathPlanetmath.


From the class equationMathworldPlanetmathPlanetmath we know the center of a finite p-group is non-trivial. Thus by inductionMathworldPlanetmath the upper central series of a p-group P terminates at P. So P is nilpotent. ∎

Example. InfiniteMathworldPlanetmath p-groups may not always be nilpotent. In the extreme there are counterexamples like the Tarski monsters Tp. These are infinite p-groups in which every proper subgroupMathworldPlanetmath has order p. Therefore given any two non-trivial elements x,y in which yx generate Tp. In particular, the only central element is 1 so that the upper central series is trivial and therefore Tp is not nilpotent.

Indeed, Tarski monsters are not in fact solvable groupsMathworldPlanetmath which is a weaker property than nilpotent.

Example. Some infinite p-groups are nilpotent. Indeed, some infinite p-groups are even abelianMathworldPlanetmath such as p – the countableMathworldPlanetmath dimensionPlanetmathPlanetmath vector spaceMathworldPlanetmath over the field p – and the Prüfer group p – the inductive limit of pn.

Theorem 2.

Let G be a finite groupMathworldPlanetmath. Then all the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    G is nilpotent.

  2. 2.

    Every Sylow subgroup of G is normal.

  3. 3.

    For every prime p||G|, there exists a unique Sylow p-subgroupMathworldPlanetmathPlanetmath of G.

  4. 4.

    G is the direct productMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of its Sylow subgroups.

For the proof recal the following consequence of the Sylow theoremsMathworldPlanetmath:

Proposition 3.

If G is a finite group and P a Sylow p-subgroup of G then


(See Subgroups Containing The NormalizersMathworldPlanetmath Of Sylow Subgroups Normalize Themselves)

Now we prove Theorem 2


(1) implies (2). Suppose that G is nilpotent and that P is a Sylow p-subgroup of G. Then as G is nilpotent, every subgroup of G is subnormal in G, meaning, if H is properly contained in G then NG(H) properly contains H. Thus NG(NG(P)) is larger than NG(P) or NG(P)=G. However because P is a Sylow p-subgroup we know NG(P)=NG(NG(P)) so we conclude NG(P)=G. Therefore every Sylow p-subgroup of G is normal in G.

(2) implies (3). Suppose every Sylow subgroup of G is normal in G. Then by the Sylow theorems we know that for every prime p dividing |G| there is exactly one Sylow p-subgroup of G – as all Sylow p-subgroups are conjugatePlanetmathPlanetmathPlanetmath and here by assumptionPlanetmathPlanetmath all are also normal.

(3) implies (4). Suppose that there is a unique Sylow p-subgroup of G for every p||G|. Then by the Sylow theorems every Sylow subgroup of G is normal in G. Furthermore, if P and Q are two distinct Sylow subgroups then they they are Sylow subgroups for different primes so that by Lagrange’s theorem their intersectionMathworldPlanetmath is trivial. Let P1,,Pk the Sylow subgroups of G. Then as each Pi is normal in G we have G=P1Pk and we have also demonstrated P1PiPi+1=1 for 2ik therefore G is the direct product of P1,,Pk.

(4) implies (1). Suppose that G is a productMathworldPlanetmathPlanetmath of its Sylow subgroups. Then as every Sylow subgroup is a p-group, G is a product of nilpotent groups so G 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