Solvable group

A solvable group has a subnormal series whose factor groups are abelian.

The assertion follows using only Sylow theorems. Write the order instead as $p{q}^2$ to separate the prime of multiplicity one. If $p<q$, the number of Sylow $q$-subgroups divides $p$ and is one modulo $q$, so one such subgroup is normal. If $p>q$ and the Sylow $p$-subgroup is not normal, its number must be $q^2$, forcing $p\mid (q-1)(q+1)$. Since $p>q$, this forces $p=q+1$ and hence $p=3,q=2$. In that order-twelve exception, four order-three subgroups account for eight nonidentity elements. The four remaining elements contain every order-four Sylow subgroup, so that subgroup is unique and normal. In every case a normal Sylow subgroup and its quotient have prime or prime-square order, hence are abelian groups. They provide a series with abelian factors and prove solubility. Orders with the squared exponent lowered to one or zero follow by the same simpler counting.

A finite group is supersolvable if it has a normal series of a group whose nontrivial factors are cyclic of prime order. This is stronger than solubility, because the individual terms must be normal in the ambient group. For instance, $A_4$ is soluble via its normal Klein four-group, but is not supersolvable: it has no normal subgroup of order two or three with which such a series could begin.

A finitely generated abelian group whose elements all have finite order is finite. Induct on derived length for a finitely generated soluble group with this property. Its abelianization is finite, so its commutator subgroup has finite index and is finitely generated by Schreier's lemma. That subgroup is soluble of smaller derived length and is still a torsion group, hence finite by induction. The ambient group is a finite extension of it and is finite.

A group is virtually solvable, or virtually soluble, when it contains a solvable group as a finite-index subgroup. This is preserved under subgroups, quotients and group extensions. The extension proof can use a finite-index characteristic soluble subgroup of the kernel, followed by centralizing the resulting finite normal kernel. Finite generation is not required for these closure properties.

Every virtually soluble group $K$ has a soluble characteristic subgroup of finite index. First take a soluble normal subgroup $K_0$ of finite index by the subgroup core construction. Choose a soluble normal subgroup $R\supseteq K_0$ maximizing its image size in the finite quotient $K/K_0$. For every soluble normal subgroup $S$, the product $RS$ is soluble, because its quotient by $R$ is a quotient of $S$. Maximality forces $S\le R$. Hence $R$ is the unique largest soluble normal subgroup and is invariant under all automorphisms. This avoids incorrectly intersecting infinitely many conjugates when proving closure under group extensions.

The derived series is defined byA group is solvable exactly when this series reaches the trivial group after finitely many steps.

For a solvable group, its derived length is the least nonnegative integer $r$ at which the derived series reaches the trivial group, with $G^{(0)}=G$ and $G^{(r+1)}$ the commutator subgroup of $G^{(r)}$. A nontrivial abelian group has derived length one; a nonabelian group with nontrivial abelian commutator subgroup has derived length two.

A group is metabelian when its commutator subgroup is abelian, equivalently when its derived length is at most two.

Let $L◃G$ and both $L,G/L$ be abelian. For an irreducible $G$-module, choose a subgroup $A\supseteq L$ of maximal order whose restriction contains a linear character $\phi$. Every subgroup containing $L$ is normal. If an element $x\notin A$ stabilizes $\phi$, its action on the nonzero $\phi$-isotypic component has an eigenvector; that line is invariant under $⟨A,x⟩$ and affords a larger linear constituent, contradicting maximality. Thus $I_G(\phi )=A$, and the Clifford correspondence gives $\chi ={\phi }^G$.

A group is polycyclic when it has a subnormal series with cyclic factors. Every subgroup and quotient of a polycyclic group is finitely generated, and extensions of polycyclic groups are polycyclic.

If a group has a cyclic-factor subnormal series of length $n$, choose one lift of a generator of each nontrivial factor. Induction gives $G_i=⟨G_{i-1},g_i⟩$, so the chosen lifts generate $G$. Thus at most $n$ generators are needed. Applying this to the intersection chain in subgroups of polycyclic groups are polycyclic bounds the number of generators of every subgroup by the same series length.

Given a cyclic-factor subnormal series $1=G_0⊴\cdots ⊴G_n=G$ and $H\le G$, intersect each term with $H$. The induced map $(H\cap G_{i+1})/(H\cap G_i)\to G_{i+1}/G_i$ is injective. Since every subgroup of a cyclic group is cyclic, the intersection chain is again a cyclic-factor subnormal series, after removing repetitions. Hence $H$ is a polycyclic group. In particular, every subgroup is finitely generated.

Every polycyclic group has a finite-index subgroup with a series whose nontrivial factors are infinite cyclic. This is the polycyclic finite-index lemma; it also supplies torsion-free finite-index subgroups. Such a subgroup is a poly-infinite cyclic group. Combining its series with the extension rule for Poincare duality groups makes it a duality group of dimension its Hirsch length.

The Hirsch length of a polycyclic group counts infinite cyclic factors in a polycyclic series. Refinement shows it is independent of the series, additive in extensions, and unchanged in finite-index subgroups. A finite-index poly-infinite cyclic group has one infinite cyclic factor for each unit of Hirsch length.

A poly-infinite cyclic group has a finite subnormal series all of whose nontrivial factors are infinite cyclic groups. Intersecting its series with a subgroup again gives trivial or infinite cyclic factors; repetitions may be deleted. Splicing a kernel series with the preimage of a quotient series proves closure under extensions. Such groups are finitely presented groups: induct on the series, split the last extension by an infinite cyclic quotient, and use finite presentation of a semidirect product.

A group is nilpotent when it has a finite central series, equivalently when its lower central series reaches the identity.

A nilpotent group is finite exactly when its abelianization is finite; finite generation is not a separate hypothesis. Induct on the nilpotency class $c$. The abelian case is immediate. Otherwise $G/{\gamma }_c(G)$ has smaller class and the same abelianization, so is finite. Let $H=G$, $K={\gamma }_{c-1}(G)$. Their commutator subgroup $[H,K]={\gamma }_c(G)$ is central. Hence its commutator map is a homomorphism in each variable. Both centralizer quotients are finite because their kernels contain ${\gamma }_c(G)$. Finitely many pairs of coset representatives therefore generate the abelian group $[H,K]$, and its exponent divides the exponent of $H/C_H(K)$. It is finite, completing the induction.

For $tv{t}^{-1}=Bv$, the commutator $[t,v]$ is $(B-I)v$ in additive notation. The lower central series from its second term is $(B-I){\mathbb{Z}}^r,(B-I{)}^2{\mathbb{Z}}^r,\dots$. It terminates exactly when $B-I$ is a nilpotent matrix. For rank two this means $(B-I{)}^2=0$, equivalently $\det B=1$ and $\mathrm{tr}B=2$. Nilpotence of a finite-index subgroup alone gives virtual nilpotence of the whole group.

Every finitely generated torsion-free group which is a nilpotent group embeds as a discrete cocompact subgroup of a unique simply connected nilpotent real Lie group. Its dimension equals the group's integral cohomological dimension of a group. In dimension three the Lie algebra is either abelian or has a basis with a single nonzero bracket $[Y,T]=X$, giving the real Heisenberg group.

If a nilpotent group is generated by $a_1,\dots ,a_d$, every element of its commutator subgroup is a product $[x_1,a_1]\cdots [x_d,a_d]$. The sets in the displayed formula need not be subgroups. Induct on the nilpotency class $c$. Collection in $G/{\gamma }_c(G)$ gives the desired product with a central error. Since ${\gamma }_c(G)=[{\gamma }_{c-1}(G),G]$, that error is a product $[y_1,a_1]\cdots [y_d,a_d]$ with $y_i\in {\gamma }_{c-1}(G)$. These factors are central, and ${\gamma }_{c-1}(G)$ centralizes $G\prime$, so replacing $x_i$ by $x_i{y}_i$ absorbs the error. For a finitely generated pro-p group, applying this formula in all finite quotients and using compactness makes its algebraically generated commutator subgroup closed.

If a nilpotent group is generated by $a_1,\dots ,a_d$, every element of its commutator subgroup is a product $[x_1,a_1]\cdots [x_d,a_d]$. The sets in the displayed formula need not be subgroups. Induct on the nilpotency class $c$. Collection in $G/{\gamma }_c(G)$ gives the desired product with a central error. Since ${\gamma }_c(G)=[{\gamma }_{c-1}(G),G]$, that error is a product $[y_1,a_1]\cdots [y_d,a_d]$ with $y_i\in {\gamma }_{c-1}(G)$. These factors are central, and ${\gamma }_{c-1}(G)$ centralizes $G\prime$, so replacing $x_i$ by $x_i{y}_i$ absorbs the error. For a finitely generated pro-p group, applying this formula in all finite quotients and using compactness makes its algebraically generated commutator subgroup closed.

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.

Choose the first term of the upper central series not contained in $H$. The preceding term lies in $H$, so an element of the new term outside $H$ commutes with $H$ modulo that preceding term and normalizes $H$.

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.

The nilpotency class of a nilpotent group $G$ is the least nonnegative integer $s$ for which ${\gamma }_{s+1}(G)=\{1\}$ in the lower central series. Thus nontrivial abelian groups have class one.

Group commutators are taken coordinatewise in a direct product of groups, giving ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$. A finite direct product of nilpotent groups is nilpotent, with class the maximum of the factor classes.

// Target: algebra.bigb

A group $G$ is two-step nilpotent whenEquivalently, its commutator subgroup is contained in the center of a group of $G$. This allows every word to be collected into powers of generators followed by powers of their pairwise commutators.

In a two-generated two-step nilpotent group, use $c=[x,y]=x^{-1}{y}^{-1}xy$. Centrality of $c$ gives $yx=xy{c}^{-1}$, so every word collects to $x^a{y}^b{c}^r$. The multiplication of these collected expressions is $(a,b,r)(a^{\prime },b^{\prime },r^{\prime })=(a+a^{\prime },b+b^{\prime },r+r^{\prime }-a^{\prime }b)$. Relations on the central commutator can reduce $r$ modulo its order. To turn collection into a word problem for a group algorithm, one must also prove the independence of the remaining coordinates, for example by constructing a group with that cocycle multiplication.

A group is virtually nilpotent when it has a nilpotent subgroup of finite index.

The lower central series of a group is defined by ${\gamma }_1(G)=G$ and ${\gamma }_{i+1}(G)=[{\gamma }_i(G),G]$. A group is nilpotent of class at most $c$ exactly when ${\gamma }_{c+1}(G)$ is trivial.

Let $G$ be a nilpotent group of class at most $s$ in which every element has order at most $r$. A subgroup generated by $k$ elements has order at mostIndeed, collection by the lower central series expresses every element as an ordered product of simple group commutators in the generators of weights at most $s$. There are at most $k+k^2+\cdots +k^s\le s{k}^s$ such commutators, and each exponent may be reduced to one of at most $r$ values.

The free nilpotent group of rank $r$ and class $c$ is $F_r/{\gamma }_{c+1}(F_r)$, where $F_r$ is the free group of rank $r$ and ${\gamma }_i$ is its lower central series.

The quotient mapis a central nonsplit extension of the Integer Heisenberg group by ${\gamma }_3(F_2)/{\gamma }_4(F_2)\cong {\mathbb{Z}}^2$. It cannot split because the source has abelianization ${\mathbb{Z}}^2$, whereas a central splitting would give an abelianization containing an additional direct factor ${\mathbb{Z}}^2$.

A central series is a chain of normal subgroups whose successive quotients lie in the centers of the corresponding quotient groups.

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