0.8 Nilpotent group

Def. Consider the upper central series $\{e\}=C_0(G)\le C_1(G)\le ...\le C_n(G),$ where $C_i(G)/C_{i-1}(G)=Z(G/C_{i-1}(G))$. If $C_n(G)=G$ for some $n$, $G$ is a nilpotent group.

Thm. $G$ is nilpotent iff $G$ is a direct product of Sylow subgroups.

Proof. Suppose $P$ is a Sylow $p$-subgroup of $G$. Since $N_G(P)=N_G(N_G(P))$, by the previous lemma $N_G(P)=P$, i.e. $P$ is normal in $G$. Thus for each prime $p$ which divides the order of $G$, Sylow $p$-subgroup is normal in $G$ and so $G$ is a direct product of Sylow subgroups. Another direction can be proved by $p$-group nilpotent and the direct product of a finite number of nilpotent groups nilpotent.

Thm. $G$ is nilpotent iff any maximal subgroup of $G$ is normal.

Proof. Assume any maximal subgroup of $G$ is normal. If $G$ is not nilpotent, then there is a Sylow subgroup $S$ such that $G>N_G(S)>S$. Then there is a maximal subgroup $M$ satisfying $S<N_G(S)<M⊲G$. By Frattini’s argument, $G=M{N}_G(S)=M$, contradiction. Conversely, if $G$ is nilpotent and $M<G$ is a maximal principle, $M<N_G(M)$ yields $N_G(M)=M$ and so $M⊲G$.