 Solution

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is