Nilpotent Groups

Upper central series, nilpotency class, and direct products of Sylow subgroups.

The idea

A nilpotent group is a group that can be built up from centers in finitely many steps. Recall that the center $Z(G)$ is the subgroup of elements that commute with every element of $G$; it equals $G$ exactly when $G$ is abelian.

To measure a nonabelian group, take centers repeatedly. Set $Z_{0} = 1$, and once $Z_{i}$ is defined, let $Z_{i+1}$ be the subgroup of $G$ whose image in $G / Z_{i}$ is the center of $G / Z_{i}$, so that $Z_{1} = Z(G)$. The chain $1 = Z_{0} \le Z_{1} \le Z_{2} \le \cdots$ is the upper central series of $G$. If $Z_{c} = G$ for some finite $c$, then $G$ is nilpotent, and the smallest such $c$ is its class. If instead some $Z_{i}$ equals $Z_{i+1}$ before the series reaches $G$ — as happens at once when $Z(G) = 1$ — the series never advances past that subgroup, and $G$ is not nilpotent.

Every finite $p$-group is nilpotent: a nontrivial finite $p$-group has a nontrivial center, so each quotient $G / Z_{i}$ contributes a new step until the series reaches $G$. Nilpotency lies strictly between abelian and solvable — every abelian group is nilpotent of class $1$, and every nilpotent group is solvable, but neither converse holds.

Ways to work on it

• Walkthrough. Climb the upper central series to measure how close a group is to abelian.
• Practice. Decide whether a finite group is nilpotent.
• Hardest. Apply the Sylow direct-product structure theorem.

Not sure where to start? Take the ten-question placement test.