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 $y\notin ⟨x⟩$ 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 $p||G|$, 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 $p||G|$. 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 $P_1,\mathrm{\dots },P_k$ 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.∎