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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 lower central series ${\gamma }_{1}G=G$, ${\gamma }_{j+1}G=[G,{\gamma }_{j}G]$ terminating at the identity. A polycyclic group has a finite subnormal series with cyclic factors, which may be finite or infinite.

If a nilpotent group is generated by finitely many elements, then ${\gamma }_{j}G/{\gamma }_{j+1}G$ is generated by the finitely many weight-$j$ iterated commutators in these generators. Modulo the next term, commutator identities make these quotients abelian and permit this collection of words. Each quotient is thus a finitely generated abelian group. Refine each by a series with cyclic factors and splice the refinements through the finite lower central series. This proves that a finitely generated nilpotent group is polycyclic.

For the duality-group assertion, first prove the useful finite-index reduction. Induct on the length of a polycyclic series, taking its last proper term $H◃G$. By induction $H$ has a finite-index poly-infinite cyclic group. Because $H$ is finitely generated, it has only finitely many subgroups of any fixed finite index. Intersect all automorphic images of that subgroup to obtain a characteristic finite-index subgroup $H_0\subset H$ which is still poly-infinite cyclic. It is normal in $G$. The quotient $G/H_0$ is finite-by-cyclic. If it is finite, $H_0$ already suffices; if its cyclic quotient is infinite, the subgroup generated by a lift of its generator is infinite cyclic and has finite index. Its preimage in $G$ is an extension of $H_0$ by $\mathbb{Z}$, hence poly-infinite cyclic. This proves polycyclic groups are virtually poly-infinite cyclic.

Take such a finite-index subgroup $P$ of the given dimension-three Poincare duality group. It is again a dimension-three duality group. The extension rule proved in Question 3, applied successively to its infinite cyclic series, makes $P$ a duality group of dimension equal to its Hirsch length. Hence that length is three. Its last quotient giveswhere $H$ has a poly-infinite cyclic series of length two. Thus $H\cong \mathbb{Z}⋊\mathbb{Z}$, with the last generator acting on the first by $+1$ or $-1$. In the first case $H={\mathbb{Z}}^2$. In the second case $H$ is the Klein-bottle group $⟨a,b\mid ba{b}^{-1}=a^{-1}⟩$. Its characteristic subgroup $⟨a,b^2⟩\cong {\mathbb{Z}}^2$ is the centralizer of its commutator subgroup $⟨a^2⟩$. Replace $H$ by this characteristic index-two subgroup if necessary. It remains normal in $P$, and the quotient is finite-by-infinite-cyclic. Taking the preimage of a finite-index infinite cyclic subgroup givesThis extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0={\mathbb{Z}}^2{⋊}_B\mathbb{Z}$ for some $B\in G{L}_2(\mathbb{Z})$.

In additive notation on ${\mathbb{Z}}^2$, $[t,v]=(B-I)v$. All its other internal commutators vanish, so induction givesConsequentlyThis is the nilpotence criterion for an abelian-by-cyclic group. If $B=I$ the group is abelian; otherwise it is two-step nilpotent. The criterion concerns this actual semidirect product. A finite extension of it is only known to be virtually nilpotent: for example ${\mathbb{Z}}^2{⋊}_{-I}\mathbb{Z}$ is virtually abelian but its lower central series never terminates.

Write elements of the real Heisenberg group as $(x,y,z)$. Direct matrix multiplication and inversion giveThese formulas show that $N_k=(k\mathbb{Z}{)}^3$ is a subgroup. It is discrete. For the right quotient, multiplying by $(ka,kb,kc)$ changes the coordinates toThus $(x\mathrm{mod}k,y\mathrm{mod}k)$ is well defined on the quotient. Over a small base rectangle choose real lifts $x,y$ of its coordinates. Every coset above that rectangle has a representative $(x,y,z)$, and two such representatives differ precisely by $z\mapsto z+kc$. Henceis a smooth local trivialization. The fibres are circles, proving the locally trivial fibration over ${\mathbb{R}}^2/k{\mathbb{Z}}^2\cong T^2$. Reducing first $x,y$ and then $z$ also gives a compact fundamental region, so $N_k$ is cocompact.

The ambient Lie group is diffeomorphic to ${\mathbb{R}}^3$, hence simply connected and contractible. Its quotient has fundamental group $N_k$, and the first integral homology is its abelianization. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^m{b}^n{c}^{\ell }$, since its third coordinate is $k^{2}mn+k\ell$. The commutator formula isso $c$ is central and $[a,b]=c^k$. Every commutator lies in $(0,0,k^2\mathbb{Z})$, and this subgroup is generated by $[a,b]$. Thus $N_k^{\prime }=⟨c^k⟩$. The homomorphism $(x,y,z)\mapsto (x/k,y/k,z/k\mathrm{mod}k)$ is surjective onto ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$ and has exactly this kernel; the cross term in multiplication is divisible by $k$ after dividing by $k$. ThereforeFor $k=1$ the last factor is trivial. This is the Scaled Heisenberg lattice, whose fibre is central but whose total group need not split as a direct product of groups.

Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a minimal normal subgroup $K/D$ of $G/D$ contained in $C/D$. In a finite soluble group, such a minimal normal subgroup is elementary abelian: the last nontrivial term of its derived series is an abelian characteristic subgroup, so minimality makes the whole subgroup abelian; its primary components and subgroup of elements killed by $p$ then force one prime and exponent $p$.

It follows that $K^{\prime }\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K^{\prime }\le Z(K)$. Therefore $K$ is a nilpotent group of class at most two. Since $K$ is normal in $G$, the definition of the Fitting subgroup gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the Fitting subgroup is self-centralizing in soluble groups:

It suffices to centralize each minimal normal subgroup $M$. The subgroup $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first caseIn the second case $M$ is a nilpotent group. Its nontrivial group center is a characteristic subgroup of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are characteristic subgroups, and the subgroup killed by a prime show that $M$ is elementary abelian of exponent $p$. Thus $M\le O_p(G)$.

The finite p-group $O_p(G)$ acts by conjugation on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne 1$. This intersection is normal in $G$, and minimality makes it $M$. The other p-cores commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the group socle is generated by these minimal normal subgroups,

Put $\Phi =\Phi (G)$ and let $H$ be the full preimage of $F(G/\Phi )$ in $G$. The preceding part gives $\Phi \le F(G)$, so the normal nilpotent image $F(G)/\Phi$ lies in $F(G/\Phi )$.

For the reverse inclusion, choose a Sylow subgroup $P$ of $H$. Its image $P\Phi /\Phi$ is the unique Sylow subgroup of the nilpotent group $H/\Phi$, hence a characteristic subgroup there and normal in $G/\Phi$. Thus $P\Phi ⊴G$. The subgroup $P$ is also a Sylow subgroup of $P\Phi$: the p-part of $\Phi$ lies in $P$, because $P\cap \Phi$ is a Sylow subgroup of the normal subgroup $\Phi$ of $H$. Apply the Frattini argument to $P\Phi$:The Frattini subgroup property forces $N_G(P)=G$. Hence every Sylow subgroup of $H$ is normal, so $H$ is a nilpotent group and $H\le F(G)$. Together the inclusions proveThis is Frattini lifting of nilpotence; simply calling an extension of nilpotent groups nilpotent would not establish it.

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.

Start with the prime decomposition of a closed orientable three-manifold. Its fundamental group is a free product of the prime-summand groups. A nontrivial abelian group cannot be a free product of two nontrivial groups. The same is true of a nontrivial finitely generated nilpotent group: it has nontrivial center, whereas such a free product has trivial center. Thus all nontrivial fundamental group comes from a single prime summand; simply connected summands do not affect the argument.

A prime but nonirreducible orientable summand is $S^2\times S^1$, giving $\Gamma \cong \mathbb{Z}$. Otherwise the infinite-group prime summand is an irreducible three-manifold. The sphere theorem for three-manifolds gives ${\pi }_2=0$. Its simply connected universal cover is noncompact, so its top ordinary homology vanishes; its first and second homology also vanish by simple connectivity and the Hurewicz theorem. It is therefore acyclic and contractible by the Whitehead theorem. The summand is aspherical, and its closed orientable three-dimensional classifying space gives integral cohomological dimension of a group equal to three: the cellular resolution gives the upper bound and top Poincare duality gives the lower bound.

If $\Gamma$ is torsion-free abelian, finite generation makes it ${\mathbb{Z}}^r$. Its integral cohomological dimension of a group is $r$, as one sees from the standard $r$-torus. The aspherical case therefore has $r=3$. We obtainBoth occur, on $S^2\times S^1$ and the three-torus.

For the infinite torsion-free nilpotent case, the same reduction gives either $\mathbb{Z}$ or an aspherical group of cohomological dimension three. By the Malcev completion of a torsion-free nilpotent group, the latter is a lattice in a simply connected three-dimensional nilpotent Lie group. A nonabelian three-dimensional nilpotent Lie algebra has abelianization of dimension at least two: a nilpotent algebra generated modulo its derived algebra by one element would be generated by that element and thus be abelian. Its derived algebra is therefore one-dimensional, and nilpotence makes that line central. Hence it has a basis with a single nonzero bracket $[Y,T]=X$; the other possibility is an abelian algebra. In the nonabelian case the lattice's center is an isolated infinite cyclic subgroup, generated by $x$, and the quotient by it is ${\mathbb{Z}}^2$. Choose lifts $y,t$ of a basis of that quotient. Their commutator is $x^b$ for a nonzero integer $b$, and normal forms giveThe subgroup $A=⟨x,y⟩\cong {\mathbb{Z}}^2$ is normal and ${\Gamma }_b/A\cong \mathbb{Z}$. In its ordered basis $x,y$, conjugation by $t$ acts byThis is the unipotent torus-bundle group. Indeed, its normal forms multiply according toProjection to the $n,k$ coordinates, followed by inspection of the central $m$ coordinate, proves torsion-freeness. All commutators lie in $⟨x^b⟩$, which is central, proving nilpotence. The mapping torus of the torus automorphism with this matrix realizes every integer $b$ as a closed orientable three-manifold; $b=0$ gives ${\mathbb{Z}}^3$, while $b\ne 0$ gives the Nil cases. Thus the complete list of groups under the nilpotent hypotheses is $\mathbb{Z}$ and the groups ${\Gamma }_b$.

Here is the center argument, with its literal qualification. If a nilpotent group with abelian quotient $\Gamma /A$ were nonabelian, the last nonzero term ${\gamma }_c\Gamma$ of its lower central series would satisfyConsequently $Z(\Gamma )\cap A=1$ forces the group to be abelian. For the specific nontrivial rank-two $A$ in the printed family the hypothesis cannot actually occur: when $b\ne 0$, $x\in Z({\Gamma }_b)\cap A$; when $b=0$, the intersection is all of $A$. The implication is valid, but should not be used to invent an example satisfying this impossible extra condition.

A torsion-free finite extension of a lattice with quotient $C_4$ is realized by the quarter-turn flat three-manifold. On ${\mathbb{R}}^3$ letLet $c$ be unit translation in the third coordinate. Then $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$, $sc{s}^{-1}=c$, and $s^4=c$. The lattice $\Lambda =⟨a,b,c⟩\cong {\mathbb{Z}}^3$ is normal and the quotient is $C_4$. A representative $a^m{b}^n{c}^k{s}^j$, $0\le j<4$, with $j\ne 0$ has third-coordinate displacement $k+j/4\ne 0$, so has no fixed point; for $j=0$ a nonidentity element is a nonzero translation. This proves torsion-freeness. The action is properly discontinuous, cocompact and orientation preserving, so it gives a closed orientable flat three-manifold with precisely the required extension.

For faithful linear holonomy on the maximal translation lattice, the holonomy groups of closed orientable flat three-manifolds areThe cyclic examples replace the quarter-turn by a planar lattice rotation of order $m=2,3,4,6$ and a translation of $1/m$ along its axis. The square and triangular plane lattices realize these orders; the trace condition $2\cos (2\pi /m)\in \mathbb{Z}$ explains the crystallographic restriction. The remaining case is obtained fromTheir squares translate by one in the first and second directions, and $(\alpha \beta {)}^2$ translates by minus one in the third. Their linear parts generate $C_2\times C_2$. The three nontranslation cosets have half-integral displacement along their respective fixed axes, so every element acts without a fixed point. These constructions realize every member of the orientable three-dimensional Euclidean holonomy classification.

The maximality/faithfulness restriction is essential. Taken literally as a question about arbitrary finite group quotients, the wording has more answers: $\mathbb{Z}$ has every finite cyclic quotient, and ${\mathbb{Z}}^3$ has, for example, quotient $C_5$. An exact criterion for the literal extension question is as follows. Choose an action $Q\to \mathrm{Aut}(A)$ and an extension class $e\in H^2(Q,A)$, where $A=\mathbb{Z}$ or ${\mathbb{Z}}^3$. For every subgroup $⟨q⟩$ of prime order $p$, requireTo prove this, lift $q$ to $g$. Its $p$th power is in $A^q$, and replacing $g$ by $ag$ changes that power by the displayed norm of $a$. Thus a lift of order $p$ exists exactly when this restriction is zero. Any torsion element has a prime-order power, so the condition is both necessary and sufficient. For an oriented flat realization in rank three the linear action must also preserve orientation. This distinguishes a complete extension criterion from the finite faithful-holonomy list.

A central series is a finite chain of normal subgroupswith $G_i/G_{i-1}\le Z(G/G_{i-1})$ for every $i\ge 1$. Equivalently, writing $[x,y]=x^{-1}{y}^{-1}xy$ and $[A,B]$ for the subgroup generated by these group commutators, the condition is $[G_i,G]\le G_{i-1}$. Centrality is relative to the whole quotient group, not just to the next subgroup in the chain.

A group is nilpotent if it admits a finite central series. The trivial group admits a series of length zero. For a nontrivial nilpotent group, the minimum possible length is its nilpotency class.

Define the lower central series by ${\Gamma }_1(G)=G$ and ${\Gamma }_{j+1}(G)=[{\Gamma }_j(G),G]$. Its terms are characteristic normal subgroups and form a descending chain. Define the upper central series by $Z_0(G)=1$ andEquivalently, $Z_{i+1}(G)=\{x\in G:[x,G]\subseteq Z_i(G)\}$. These are characteristic normal subgroups in an ascending chain, and successive factors are central in the appropriate quotient.

For any length-$r$ central series $1=G_0\le \cdots \le G_r=G$, prove the two comparisons independently. Starting at ${\Gamma }_1(G)=G_r$, if ${\Gamma }_{r-i+1}(G)\le G_i$, thenDescending induction gives the lower containment. Starting at $G_0=Z_0(G)=1$, if $G_{i-1}\le Z_{i-1}(G)$, then $[G_i,G]\le G_{i-1}\le Z_{i-1}(G)$, which by definition gives $G_i\le Z_i(G)$. Thus the central-series comparison theorem yieldsIn particular, a nilpotent group satisfies ${\Gamma }_{r+1}(G)=1$ and $Z_r(G)=G$.

Conversely, if ${\Gamma }_n(G)=1$, reverse the finite lower chain to obtain $1={\Gamma }_n\le {\Gamma }_{n-1}\le \cdots \le {\Gamma }_1=G$. The defining commutator recurrence makes it a central series. If $Z_n(G)=G$, the finite upper chain is already a central series. This provesFinally, for an integer $c>0$, termination at ${\Gamma }_{c+1}=1$ gives a central series of length $c$ and hence $Z_c=G$ by the comparison. Conversely, $Z_c=G$ gives a central series of length $c$ and hence ${\Gamma }_{c+1}=1$. ThereforeRepeated terms cause no problem; the least such $c$ for a nontrivial group is its nilpotency class.

For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.