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 (http://planetmath.org/Finite) $p$ -groups are nilpotent.

Proof.

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

Example. Infinite $p$ -groups may not always be nilpotent. In the extreme there are counterexamples like the Tarski monsters $T_{p}$ . These are infinite $p$ -groups in which every proper subgroup has order $p$ . Therefore given any two non-trivial elements $x, y$ in which $y\notin\langle x\rangle$ generate $T_{p}$ . In particular, the only central element is 1 so that the upper central series is trivial and therefore $T_{p}$ is not nilpotent.

Indeed, Tarski monsters are not in fact solvable groups which is a weaker property than nilpotent. $\Box$

Example. Some infinite $p$ -groups are nilpotent. Indeed, some infinite $p$ -groups are even abelian such as $\mathbb{Z}_{p}^{\infty}$ – the countable dimension vector space over the field $\mathbb{Z}_{p}$ – and the Prüfer group $\mathbb{Z}_{p^{\infty}}$ – the inductive limit of $\mathbb{Z}_{p^{n}}$ . $\Box$

Theorem 2.

Let $G$ be a finite group. Then all the following are equivalent.

1.

$G$ is nilpotent.
2.

Every Sylow subgroup of $G$ is normal.
3.

For every prime $p\big{|}|G|$ , there exists a unique Sylow $p$ -subgroup of $G$ .
4.

$G$ is the direct product of its Sylow subgroups.

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

Proposition 3.

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

N_{G}(N_{G}(P))=N_{G}(P).

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

Now we prove Theorem 2

Proof.

(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 $N_{G}(H)$ properly contains $H$ . Thus $N_{G}(N_{G}(P))$ is larger than $N_{G}(P)$ or $N_{G}(P)=G$ . However because $P$ is a Sylow $p$ -subgroup we know $N_{G}(P)=N_{G}(N_{G}(P))$ so we conclude $N_{G}(P)=G$ . 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 conjugate and here by assumption all are also normal.

(3) implies (4). Suppose that there is a unique Sylow $p$ -subgroup of $G$ for every $p\big{|}|G|$ . 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 intersection is trivial. Let $P_{1},\dots,P_{k}$ the Sylow subgroups of $G$ . Then as each $P_{i}$ is normal in $G$ we have $G=P_{1}\cdots P_{k}$ and we have also demonstrated $P_{1}\cdots P_{i}\cap P_{i+1}=1$ for $2\leq i\leq k$ therefore $G$ is the direct product of $P_{1},\dots,P_{k}$ .

(4) implies (1). Suppose that $G$ is a product 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