Nilpotent Group

lower central series

We say $G$ is a nilpotent group if it has a lower central series terminating in the trivial subgroup after finitely many steps. That is, a series of normal subgroups

$$G=G_0⊳G_1⊳\cdots ⊳G_n=\{1\}$$

where $G_{i+1}=[G_i,G]$. The length of the lower central series $n$ is called the nilpotency class of $G$.

central series

A group $G$ is called nilpotent if it has a chain of subgroups, which is called a central series,

$$G=K_0>K_1>\cdots >K_n=1$$

such that $K_i⊲G$ and $K_i/K_{i+1}⩽Z(G/K_{i+1})$.

upper central series

A group $G$ has an upper central series terminating in the whole group after finitely many steps. That is, a series of normal subgroups

$$\{1\}=Z_0◃Z_1◃\cdots ◃Z_n=G$$

where $Z_1=Z(G)$ and $Z_{i+1}$ is the subgroup such that $Z_{i+1}/Z_i=Z\left(G/Z_i\right)$.

nilpotency class

For a nilpotent group, the smallest $n$ such that $G$ has a central series of length $n$ is called the nilpotency class of $G$; and $G$ is said to be nilpotent of class $n$.

Equivalently, the nilpotency class of $G$ equals the length of the lower central series or upper central series.

Observations.

• A nilpotent group is a solvable group.
• For any prime $p$, a $p$-group is a nilpotent group.
• Direct product of two nilpotent groups is nilpotent.
• $A_4$ and $S_4$ are not nilpotent.
• Note that $Z_{n-i}⩾K_i⩾G_i$. (Maybe that’s why they are called upper/lower central series.)

Theorem

The followings are equivalent:

• $G$ is nilpotent.
• if $H<G$, then $N_G(H)>H$.
• each maximal subgroup of $G$ is normal and of prime index.
• $G$ is a direct product of its Sylow subgroups.

\begin{proof} Assume (i) holds. Then $G$ has a central series

$$G=K_1>K_2>\cdots >K_{s+1}=1.$$

Let $H<G$. Then there exists $i$ such that $H⩾K_{i+1}$ and $H\ngeqq K_i$. It yields that $[H/K_{i+1},Z(G/K_{i+1})]=1$. So for any $h\in H$ and $y\in K_i$, there is $(h{K}_{i+1})(y{K}_{i+1})=(y{K}_{i+1})(h{K}_{i+1})$ and so $hy=(y{c}_1)(h{c}_2)$ for some $c_1,c_2\in K_{i+1}$. We have

$$(y{c}_1)(h{c}_2)=yh{c}_1^h{c}_2=yh{c}_1^{\prime }{c}_2\in yh{K}_{i+1}.$$

So $y^{-1}hy=h{c}_1^{\prime }{c}_2\in H{K}_{i+1}=H$, i.e. $y\in N_G(H)$ and $K_i⩽N_G(H)$. Thus $N_G(H)⩾H{K}_i>H$, as in (ii).

Assume (ii) holds. Let $M$ be a maximal subgroup of $G$. Then $M⊲N_G(M)⩽G$ and so $M⊲G$. The factor group $G/M$ is simple as $M$ is maximal. So $|G/M|$ is a prime, as in (iii).

Assume (iii) holds. Let $P\in {\mathrm{Syl}}_p(G)$. If $P$ is not normal in $G$, then $P<G$ and $P<N_G(P)<G$. Let $M$ be a maximal subgroup of $G$ such that $N_G(P)⩽M<G$. Then $M⊲G$. By Frattini’s argument, $G=N_G(P)M=M$, a contradiction. Hence, $P⊲G$ and so $G$ is a direct product of Sylow subgroups.

Assume (iv) holds. Since $p$-group is nilpotent and direct product of nilpotent groups is nilpotent, we obtain (i). \end{proof}

Remark. Note that $N_G(N_G(P))=N_G(P)$ for any Sylow $p$-subgroup. So (ii)→(iv) is easy.

Proposition

Let $N_1,N_2⊲G$ be nilpotent groups. Then $N_1{N}_2$ is also a normal nilpotent group.

\begin{proof} \end{proof}