PlanetMath: 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.

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 $\mathbb{Z}_{p^n}$ . $\Box$

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 $P_1\cdots P_i\intersect P_{i+1}=1$ 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. $\qedsymbol$