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For a length-$r$ central series $1=G_0\le \cdots \le G_r=G$, induction from the top gives ${\Gamma }_{r-i+1}(G)\le G_i$, and induction from the bottom gives $G_i\le Z_i(G)$. Consequently the lower central series terminates at ${\Gamma }_{c+1}=1$ exactly when the upper central series reaches $Z_c=G$.

// Target: algebra.bigb

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 soluble group is a group whose derived series $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$ terminates at $1$. Equivalently, it has a finite subnormal series with abelian factors. For the reverse implication, if $G/N$ has derived length $r$, then $G^{(r)}\le N$; if $N$ has derived length $s$, then $G^{(r+s)}=1$. Iterating this observation through the series proves the equivalence. For a finite group, refining the abelian factors to composition factors says that solubility is equivalent to all composition factors being cyclic of prime order.

Subgroups and quotient groups of soluble groups are soluble, since their derived series lie in, or are images of, the original series. The same calculation proves closure under group extensions. Finite p-groups are nilpotent groups, hence soluble: their nontrivial centers allow induction to construct a central series. Every finite nilpotent group is a direct product of groups of its Sylow subgroups. The converse fails: $S_3$ has the abelian-factor series $1<C_3<S_3$, but its order-two Sylow subgroups are not normal. The symmetric group $S_4$ is also soluble, with series $1<V_4<A_4<S_4$ and abelian factors. A finite soluble group has elementary abelian minimal normal subgroups, by the argument in 1(c). In contrast, a nonabelian simple group is a perfect group and is not soluble.

The Fitting subgroup collects the normal nilpotent structure. For a nontrivial finite soluble group it is nontrivial, since it contains an elementary abelian minimal normal subgroup. Part 1(c) proves $C_G(F)\le F$, so conjugation embeds $G/C_G(F)$ in $\mathrm{Aut}(F)$. This shows why controlling a soluble group through normal abelian or nilpotent layers is effective, even when the whole group is not a nilpotent group.

For a set of primes $\pi$, a Hall subgroup of type $\pi$ has order divisible only by primes in $\pi$ and index divisible only by primes outside $\pi$. The Hall theorem for soluble groups says that a finite soluble group has such subgroups, any two are conjugate, and every subgroup of $\pi$-order is contained in one. These assertions generalize the Sylow theorems, which treat a single prime. Here is a proof of all three assertions.

First we need coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary abelian p-group and $Q=E/V$ has order $h$ prime to $p$. A section $s:Q\to E$ defines an action on $V$ and an additive factor set $f(x,y)$ by $s(x)s(y)=f(x,y)s(xy)$. Associativity saysSumming over $z\in Q$ and dividing by $h$ in $V$ gives $f(x,y)=b(x)+xb(y)-b(xy)$, where $b(x)=h^{-1}{\sum }_{z}f(x,z)$. Replacing $s(x)$ by $-b(x)s(x)$ removes the factor set, giving a complement. Two complements differ by a map $d$ satisfying $d(xy)=d(x)+xd(y)$. Averaging over $y$ gives $d(x)=b-xb$, so the complements are conjugate by an element of $V$. This proves both splitting and conjugacy, without assuming that an arbitrary extension splits.

Induct on $|G|$, taking an elementary abelian minimal normal subgroup $V$ of exponent $p$. Choose a quotient Hall subgroup $\bar{H}\le G/V$ and let $K$ be its preimage. If $p\in \pi$, then $K$ itself is a Hall subgroup. If $p\notin \pi$, the preceding splitting argument gives a complement $H$ to $V$ in $K$, and that complement is a Hall subgroup of $G$.

For conjugacy, the images of two Hall subgroups are quotient Hall subgroups, so induction allows them to have the same image and lie in the same $K$. If $p\in \pi$, every Hall subgroup contains $V$: its product with $V$ is a $\pi$-subgroup and cannot be larger by the subgroup-order formula. Both are then $K$. If $p\notin \pi$, both are complements to $V$ in $K$, so the averaging argument conjugates them.

Finally let $U\le G$ be any $\pi$-subgroup. By induction its image lies in a quotient Hall subgroup, so $U\le K$. If $p\in \pi$, $K$ is already a Hall subgroup containing $U$. Otherwise write $K=V⋊H$. In $VU$, both $U$ and the inverse image in $H$ of the image of $U$ are complements to $V$. Complement conjugacy therefore places $U$ in a conjugate of $H$. This proves embedding and completes the induction.

Thus existence, conjugacy and containment all hold for every prime set in a finite soluble group. The solubility hypothesis matters: $A_5$ has no Hall subgroup for $\pi =\{2,5\}$. Such a subgroup would have index three; its coset action would give a nontrivial group homomorphism $A_5\to S_3$, impossible because $A_5$ is a simple group and has order $60$. This example also distinguishes a Hall subgroup from a merely arbitrary subgroup whose order uses the chosen primes.

A subnormal series of a finite group is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$. A normal series of a group requires every $G_i$ normal in the whole group. This distinction matters: a composition series is a subnormal series with nontrivial simple factors, whereas a chief series is a strict, unrefinable normal series of a group, whose factors are minimal normal subgroups of the appropriate ambient quotients. Both exist in a finite group, by successively choosing maximal proper normal subgroups or a maximal chain of ambient-normal subgroups. The direct-product result proved above shows that chief factors are powers of isomorphic simple groups, rather than necessarily simple themselves.

The Jordan–Hölder theorem says that any two composition series have the same multiset of simple factors, up to isomorphism and order. Here is an induction proof. The assertion is trivial for the identity group. Let $A,B$ be the penultimate terms of two series for $G$; they are maximal proper normal subgroups. If $A=B$, apply induction there. Otherwise $AB=G$, since its image is a nontrivial normal subgroup in each simple quotient. Put $C=A\cap B$. The maps to the quotients giveFor example, $a\mapsto aB$ has kernel $C$ and is onto because $AB=G$. Take a composition series of $C$ and append $A$, obtaining a composition series of $A$ because $A/C$ is simple; do the same with $B$. Induction inside these smaller groups shows that the factors in the two original series are, in both cases, the factors of $C$ together with $G/A$ and $G/B$. This proves the theorem and justifies the phrase Jordan–Hölder factors independently of a chosen series.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic subgroups, hence normal. A group is soluble if this series reaches the identity. Equivalently it admits a subnormal series with abelian factors. One direction uses the derived series itself. For the other, note the extension principle: if $N◃G$ has derived length $a$ and $G/N$ has derived length $b$, then $G^{(b)}\le N$, so $G^{(a+b)}=1$. Applying this principle along an abelian-factor series proves solubility.

Subgroups and quotients of a soluble group are soluble: $H^{(i)}\le G^{(i)}$ for a subgroup, and the commutator subgroup of a quotient is the image of the commutator subgroup. A finite soluble simple group is abelian, since its commutator subgroup is normal and cannot equal the whole group if the derived series terminates. An abelian simple finite group is cyclic of prime order. Therefore a finite group is soluble exactly when all its Jordan–Hölder factors are cyclic of prime order: necessity follows from the subgroup/quotient property, and sufficiency follows from the extension principle. Its chief factors are elementary abelian, by the minimal-normal-subgroup result.

For nilpotence, the lower central series is ${\gamma }_1(G)=G$, ${\gamma }_{i+1}(G)=[{\gamma }_i(G),G]$. A group is nilpotent if ${\gamma }_{c+1}=1$ for some $c$. An ascending central series $1=N_0\le \cdots \le N_c=G$ satisfies $[N_{i+1},G]\le N_i$. Reversing a terminating lower central series gives such a central series. Conversely, induction gives ${\gamma }_i(G)\le N_{c-i+1}$, so any such central series forces ${\gamma }_{c+1}=1$. The upper central series $Z_0=1$, $Z_{i+1}/Z_i=Z(G/Z_i)$ contains every ascending central series term by term, and thus reaches $G$ exactly in the nilpotent case. Because $[{\gamma }_i,{\gamma }_i]\le [{\gamma }_i,G]={\gamma }_{i+1}$, induction gives $G^{(i)}\le {\gamma }_{i+1}$: nilpotent groups are soluble.

For finite groups there is the stronger structural characterization: a finite group is nilpotent exactly when it is the direct product of groups formed from its Sylow subgroups. To prove the forward direction, the upper central series gives the normalizer condition. If $H<G$, choose its first term $Z_i$ not contained in $H$, with $Z_{i-1}\le H$. An element of $Z_i\setminus H$ has all group commutators with $H$ in $Z_{i-1}$, so normalizes $H$. Hence $H<N_G(H)$.

For a Sylow subgroup $P$, its normalizer is self-normalizing. Indeed, if $x$ normalizes $N_G(P)$, the two Sylow subgroups $P^x,P$ of that normalizer are conjugate there; adjusting $x$ by an element of $N_G(P)$ puts it in $N_G(P)$ already. Applying the strict normalizer condition to $N_G(P)$ shows that it cannot be proper, so $P$ is normal. Normal Sylow subgroups of distinct primes commute, since their group commutators lie in their trivial intersection. Their product has cardinality $|G|$ and is direct.

Conversely, a nontrivial finite $p$-group has nontrivial centre: in the class equation every noncentral conjugacy class has size divisible by $p$, so $p$ divides the centre's order. Induct on its order through the quotient by its centre to obtain an upper central series. Thus every finite $p$-group is nilpotent, and a finite direct product of groups of them is nilpotent because its upper central series is obtained componentwise.

A finite supersolvable group has a normal series of a group with cyclic prime-order factors. It is soluble, but the ambient-normal condition is stronger than the condition on a composition series. Finite nilpotent groups are supersolvable: a $p$-group has a central subgroup of order $p$, and induction in its quotient builds an ambient-normal prime-factor series; direct products of groups of the Sylow factors combine these series. The converses fail. The group $S_3$ is supersolvable via $1<C_3<S_3$, but is not nilpotent because its Sylow $2$-subgroups are not normal. The group $A_4$ is soluble via $1<V_4<A_4$, but is not supersolvable: no subgroup of order two is normal, since the three involutions are conjugate, and no subgroup of order three is normal, since there are four such subgroups. In this example $1<C_2<V_4<A_4$ is a composition series whose first $C_2$ is not normal in $A_4$, while $1<V_4<A_4$ is a chief series with a nonsimple chief factor $V_4$.

Finally, a perfect group satisfies $G^{\prime }=G$. In any finite group, the derived series eventually stabilizes at a perfect characteristic subgroup, because a descending sequence of finite subgroup orders eventually stops decreasing. It is trivial exactly when the group is soluble. A nonabelian simple group is perfect: its nontrivial commutator subgroup is normal and must be the whole group. Thus series distinguish successively stronger notions, summarize the building blocks through Jordan–Hölder factors, and separate soluble behaviour from a persistent perfect part.