= Solution
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 $1=G_0\leq G_1\leq\cdots\leq G_c=G$ with $[G_i,G]\leq G_{i-1}$. 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 $3\times3$ <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 $C_3\triangleleft S_3$ 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 $Z_{i-1}\leq H$. 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, $[P,Q]\leq P\cap Q=1$, 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 gives
$$
\boxed{G\cong\prod_{p\mid |G|}P_p.}
$$
This 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.
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