Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 / 1

For subgroups $A,B$ write $[A,B]$ for the subgroup generated by their group commutators. A nilpotent group has a central series: a finite chain $1=G_0\le G_1\le \cdots \le G_c=G$ with $[G_i,G]\le G_{i-1}$. Equivalently its upper central series, defined by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, reaches $G$. Indeed every central series term $G_i$ lies in $Z_i$ by induction, and the upper central series itself is central. A third equivalent definition is that the lower central series ${\gamma }_1=G$, ${\gamma }_{i+1}=[{\gamma }_i,G]$ terminates at ${\gamma }_{c+1}=1$: reversing this series gives a central series, and induction along any central series gives ${\gamma }_{c+1}=1$.

The least such $c$ is the nilpotency class. Nontrivial abelian groups have class one. Upper-unitriangular $3\times 3$ matrices have class two: their group commutators lie in the central top-right-entry subgroup, and further group commutators vanish. The same example works over a finite field or over the integers, so nilpotence is not a finiteness property. Subgroups and quotients inherit nilpotence by intersecting or mapping a central series; finite direct products of groups have class at most the largest factor class. Every nilpotent group is soluble, since its derived series is contained successively in the lower central series. The converse fails: $S_3$ is soluble but its center is trivial, so it cannot be a nontrivial nilpotent group. Extensions of nilpotent groups need not be nilpotent, as the extension $C_3◃S_3$ with quotient $C_2$ shows.

To prove the finite classification, first note that every nontrivial finite p-group has nontrivial center. In its class equation, every noncentral conjugacy class has size divisible by $p$, so the center's order is divisible by $p$. Induct on the group order: the smaller p-group $P/Z(P)$ is nilpotent, and lifting its central series and adjoining $Z(P)$ proves $P$ nilpotent. A finite direct product of groups of prime-power-order groups is consequently nilpotent.

Conversely suppose $G$ is finite and nilpotent. We need the normalizer condition for nilpotent groups. For any proper subgroup $H<G$, choose the first upper-central term $Z_i$ not contained in $H$. Then $Z_{i-1}\le H$. Every $z\in Z_i$ commutes with $H$ modulo $Z_{i-1}$, so it normalizes $H$. Choosing $z\notin H$ proves $H<N_G(H)$.

Let $P$ be a Sylow subgroup and put $M=N_G(P)$. Since $P$ is normal in $M$, it is the unique Sylow p-subgroup of $M$. Any element normalizing $M$ must therefore normalize $P$, so $N_G(M)=M$. If $M$ were proper, the normalizer condition would contradict this equality. Hence $N_G(P)=G$: every Sylow subgroup is normal. For Sylow subgroups $P,Q$ belonging to different primes, $[P,Q]\le P\cap Q=1$, so they commute. Their product has order the product of their orders, namely $|G|$, and has trivial intersections between each factor and the product of the others. Multiplication therefore givesThis proves finite nilpotent group decomposition in both directions without assuming the classification in the normalizer argument. It also shows that a finite group is nilpotent exactly when all its Sylow subgroups are normal. The finite hypothesis is essential to the prime-power direct-product statement; an infinite cyclic group is nilpotent but is not such a finite product.