# 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.

###### 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.

1. 1.

$G$ is nilpotent.

2. 2.

Every Sylow subgroup of $G$ is normal.

3. 3.

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

4. 4.
###### Proposition 3.

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

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

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 FiniteNilpotentGroups 2013-03-22 15:46:40 2013-03-22 15:46:40 Algeboy (12884) Algeboy (12884) 18 Algeboy (12884) Topic msc 20D15 msc 20E15 msc 20E34 NilpotentGroup DirectProductAndRestrictedDirectProductOfGroups ClassificationOfFiniteNilpotentGroups