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A finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. The normalizer condition for nilpotent groups and Sylow conjugacy force each Sylow normalizer to be the whole group. Distinct normal Sylow subgroups commute because their commutators lie in their trivial intersection. Conversely finite p-groups are nilpotent by the nontrivial center and induction; direct products preserve nilpotence.

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.

A subnormal series is a finite chain $1=G_r⊴G_{r-1}⊴\cdots ⊴G_0=G$, in which each term is normal in the preceding term. Its factors are the quotient groups $G_i/G_{i+1}$. A normal series of a group imposes the stronger condition that every term be normal in $G$. These conditions should not be conflated: inserting a subgroup normal in an intermediate term need not preserve normality in the whole group. A refinement inserts additional terms, and two series are equivalent when their nontrivial factors can be paired by group isomorphisms.

The basic refinement mechanism is the Zassenhaus butterfly lemma. For $A^{\prime }⊴A$ and $B^{\prime }⊴B$, let $C=A\cap B$ and $D=(A^{\prime }\cap B)(A\cap B^{\prime })$. Both factors of $D$ are normal in $C$, so $D⊴C$. The map from $C$ onto $A^{\prime }C/A^{\prime }(A\cap B^{\prime })$ has kernel $D$: if an element of $C$ equals $a^{\prime }{b}^{\prime }$ with $a^{\prime }\in A^{\prime }$ and $b^{\prime }\in A\cap B^{\prime }$, then $a^{\prime }=c{b}^{\prime -1}\in A^{\prime }\cap B$. Conversely each product in $D$ is in the kernel. The first isomorphism theorem therefore givesThe denominators are normal in their numerators: conjugation by $C$ preserves them, while conjugation by $A^{\prime }$ or $B^{\prime }$ changes their elements only by a factor already in that normal subgroup.

Given two subnormal series $(G_i)$ and $(H_j)$, refine the first by $G_{i+1}(G_i\cap H_j)$, and the second by $H_{j+1}(H_j\cap G_i)$. The endpoints are $G_i,G_{i+1}$ and $H_j,H_{j+1}$ respectively, and the Zassenhaus butterfly lemma pairs the successive factors indexed by $(i,j)$ and $(j,i)$. This proves the Schreier refinement theorem.

A composition series is a strict subnormal series all of whose factors are simple groups. Every finite group has one: choose a maximal proper normal subgroup, repeat inside it, and use decreasing orders to terminate. A simple factor cannot admit a nontrivial refinement. Applying the preceding refinement theorem to two composition series proves the Jordan–Hölder theorem: their lengths and multisets of simple factors agree, up to group isomorphism. The factors are the Jordan–Hölder factors, and their number is the composition length. If $N⊴G$, lifting a composition series of $G/N$ and adjoining one for $N$ gives

These factors do not determine the whole group, because they do not determine how its group extensions are assembled. For example, $C_4$ and the Klein four-group both have two factors $C_2$, but one contains an element of order four and the other does not. The symmetric group $S_3$ has a composition series $1<C_3<S_3$, with factors $C_3,C_2$, whereas the simple group $A_5$ has the single factor $A_5$.

A chief series is a maximal strict normal series of a group; each chief factor is a minimal nontrivial normal subgroup in the appropriate quotient of $G$. Finite chief factors are characteristically simple. To justify their structure, let $N$ be any nontrivial finite characteristically simple group and choose a minimal nontrivial normal subgroup $M$ of $N$. Its distinct images under automorphisms of $N$ are minimal normal subgroups, so distinct ones intersect trivially and commute. Choose a maximal family giving an internal direct product of groups. If any other image were not contained in that product, its intersection with the product would be trivial by minimality, and it could be added, a contradiction. Hence this product is characteristic and equals $N$. Each factor $M$ is simple: a proper nontrivial normal subgroup of $M$ would be normal in the whole direct product $N$, contrary to the choice of $M$. Thus $N$ is a direct product of groups of isomorphic finite simple groups. This applies to chief factors by conjugation from $G$.

Chief and composition factors need not coincide. In $A_4$, the chain $1<V_4<A_4$ is a chief series, since conjugation cycles the three order-two subgroups of $V_4$. A composition series instead inserts any one of those subgroups, giving factors $C_2,C_2,C_3$. Its inserted order-two subgroup is normal in $V_4$ but not in $A_4$. Chief series retain the ambient group's conjugation information that composition series discard.

The derived series gives a canonical criterion for a group to be soluble: put $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic, and each factor is Abelian. A finite group is soluble exactly when this series reaches the identity, equivalently when all its composition factors are cyclic of prime order. For one direction, the derived series supplies Abelian factors, whose simple refinements are cyclic of prime order. Conversely an extension of two soluble groups is soluble: if $(G/N{)}^{(r)}=1$, then $G^{(r)}\subseteq N$; if $N^{(s)}=1$, then $G^{(r+s)}=1$. Induction up a composition series with cyclic prime factors proves the other direction. In a finite soluble group, each minimal normal subgroup is elementary Abelian: its characteristic commutator subgroup must be trivial, its primary components force a single prime, and its characteristic subgroup of $p$th powers must be trivial. Thus every chief factor is a vector space over some ${\mathbb{F}}_p$. In a general finite group the Jordan–Hölder factors can instead be nonabelian simple groups.

The central series demand more than Abelian factors: each factor must be central in the corresponding quotient of the whole group. Define the upper central series by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, and the lower central series by ${\gamma }_1=G$ and ${\gamma }_{i+1}=[{\gamma }_i,G]$. A group is nilpotent exactly when the upper series reaches $G$, equivalently when the lower series reaches the identity. Indeed, for any ascending central series $1=C_0\le \cdots \le C_c=G$, induction gives $C_i\subseteq Z_i$ and ${\gamma }_{c+1-i}\subseteq C_i$; reversing the lower series gives an ascending central series when it terminates. The least such $c$ is the nilpotency class. A central series has Abelian factors, so every nilpotent group is soluble; the converse fails, as $S_3$ is soluble but has trivial group center.

For a finite p-group, the class equation shows that the group center is nontrivial, since every noncentral conjugacy class has size divisible by $p$. Applying the same argument to successive central quotients makes the upper central series reach the group. More generally a finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. Here is the nontrivial direction. A central series gives the normalizer condition for nilpotent groups: for a proper subgroup $H$, choose the first $Z_i$ not contained in $H$. Then $[Z_i,H]\subseteq Z_{i-1}\subseteq H$, so $Z_i\subseteq N_G(H)$ and $N_G(H)>H$. If a Sylow subgroup $P$ had proper normalizer $N$, then $N_G(N)>N$; but $P$ is the unique Sylow subgroup of $N$, hence characteristic there, and $N_G(N)\subseteq N_G(P)=N$, a contradiction. Thus all Sylow subgroups are normal, different ones commute because their intersection is trivial, and they form an internal direct product. Conversely each finite Sylow subgroup is nilpotent, and products of their central series give a central series for their direct product.

Composition series record simple building blocks, chief series record ambient normal structure, derived series detect solubility, and central series detect nilpotence. Their associated factors and extension data explain why the same group can have very different useful series.

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

Repeatedly take the center of a group after quotienting by the preceding term. A group is a nilpotent group exactly when a finite term is the whole group. This yields the normalizer condition for nilpotent groups.