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 nontrivial. 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 nontrivial elements $x,y$ in which $y\notin \u27e8x\u27e9$ 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. $\mathrm{\square}$
Example. Some infinite $p$groups are nilpotent. Indeed, some infinite $p$groups are even abelian^{} such as ${\mathbb{Z}}_{p}^{\mathrm{\infty}}$ – the countable^{} dimension^{} vector space^{} over the field ${\mathbb{Z}}_{p}$ – and the Prüfer group ${\mathbb{Z}}_{{p}^{\mathrm{\infty}}}$ – the inductive limit of ${\mathbb{Z}}_{{p}^{n}}$. $\mathrm{\square}$
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 $pG$, 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 $pG$. 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 ${P}_{1},\mathrm{\dots},{P}_{k}$ the Sylow subgroups of $G$. Then as each ${P}_{i}$ is normal in $G$ we have $G={P}_{1}\mathrm{\cdots}{P}_{k}$ and we have also demonstrated ${P}_{1}\mathrm{\cdots}{P}_{i}\cap {P}_{i+1}=1$ for $2\le i\le k$ therefore $G$ is the direct product of ${P}_{1},\mathrm{\dots},{P}_{k}$.
(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  20130322 15:46:40 
Last modified on  20130322 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 