Nilpotent Groups
Table of Contents

Nilpotent Groups

Definition: Let $G$ be a group and let $1$ denote the identity in $G$. The Upper Ascending Central Series of $G$ is the subnormal series $\{ 1 \} = Z_0 \triangleleft Z_1 \triangleleft ... \triangleleft Z_k \triangleleft ... \triangleleft G$ where $Z_1 = Z(G)$ and all other groups $Z_{i+1}$ are defined such that $Z_{i+1}/Z_i = Z(G/Z_i)$.

//Recall that if $G$ is a group then $Z(G)$ denotes the center of $G$, which is the set of all points in $G$ that commute with every other element in $G$, i.e.:

(1)
\begin{align} \quad Z(G) = \{ g \in G : gh = hg, \forall h \in G \} \end{align}

Recall also that a group $G$ is abelian if and only if $G = Z(G)$, that is, every element in $G$ commutes with every element in $G$.

Definition: Let $G$ be a group. If for some least $n \in \mathbb{N}$, the upper ascending central series of $G$ terminates, i.e., $Z_n = G$, that $G$ is said to be a Nilpotent Group of Class $n$.
Theorem 1: Let $G$ be a group. Then $G$ is a nilpotent group of class $1$ if and only if $G$ is an abelian group.
  • Proof: $\Rightarrow$ Suppose that $G$ is a nilpotent group of class $1$. Then the upper ascending central series of $G$ is:
(2)
\begin{align} \quad \{ 1 \} = Z_0 \trianglelefteq Z_1 = G \end{align}
  • And by definition, we have that $Z_1/Z_0 = Z(G)$. That is, $G = Z(G)$. So $G$ is an abelian group.
  • $\Leftarrow$ Suppose that $G$ is an abelian group. Then $G = Z(G)$. So $\{ 1 \} \trianglelefteq G$ is an upper ascending central series of $G$ since $Z_1/Z_0 = G = Z(G)$. So $G$ is a nilpotent group of class $1$. $\blacksquare$
Theorem 2: Let $G$ be a group. If $G$ is a nilpotent group of class $c$ for any $c \in \mathbb{N}$ then $G$ is a solvable group.
  • Proof: Let $G$ be a nilpotent group of class $c$ and consider the following ascending central series of $G$:
(3)
\begin{align} \quad \{ 1 \} = Z_0 \triangleleft Z_1 \triangleleft ... \triangleleft Z_c = G \end{align}
  • Consider the factor $Z_{i+1}/Z_i = Z(G/Z_i)$. Then every factor if the center of some group. But the center of a group is an abelian group. So each factor $Z_{i+1}/Z_i$ is an abelian group. Hence $G$ is a solvable group. $\blacksquare$
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