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A soluble group, also called a solvable group, has a terminating derived series:For a subgroup $S\le G$, induction gives $S^{(j)}\le G^{(j)}$, so subgroups of soluble groups are soluble. For a surjective group homomorphism $q:G\to Q$, $q(G^{(j)})=Q^{(j)}$, so quotients of soluble groups are soluble. Finally, in a group extension $1\to K\to E\to Q\to 1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\le K$ and $E^{(s+r)}=1$. Soluble groups are closed under subgroups, quotients and group extensions.

A virtually soluble group contains a soluble group as a finite-index subgroup. The finite-index facts proved in parts (i)–(iii) imply closure under subgroups and quotients: intersect a finite-index soluble subgroup with the chosen subgroup, or take its image under the quotient map.

For group extensions, no finite-generation hypothesis may be inserted. We first establish the finite-index characteristic soluble subgroup lemma. If $K$ is a virtually soluble group, the kernel of its action on the cosets of a finite-index soluble subgroup is a soluble normal subgroup $K_0$ of finite index. Among soluble normal subgroups containing $K_0$, choose $R$ with maximal $|R/K_0|$, possible because $K/K_0$ is finite. If $S$ is any soluble normal subgroup of $K$, then $RS$ is soluble: it is an extension of $R$ by $S/(R\cap S)$. Maximality forces $S\le R$. Thus $R$ is the unique largest soluble normal subgroup of $K$, making it a characteristic subgroup, and it has finite index.

Now suppose $1\to K\to G\stackrel{\pi }{\to }Q\to 1$ has both $K$ and $Q$ virtually soluble. Replace $G$ by the preimage $G_0$ of a finite-index soluble subgroup of $Q$. The subgroup $R$ just constructed is characteristic in $K$ and therefore normal in $G_0$. In $E=G_0/R$, the subgroup $A=K/R$ is a finite normal subgroup, and $E/A$ is a soluble group. The centralizer $C_E(A)$ has finite index in $E$, since conjugation gives a map $E\to \mathrm{Aut}(A)$ with finite image. Its intersection with $A$ is the center of a group $Z(A)$, an abelian group, while its quotient by $Z(A)$ embeds in the soluble group $E/A$. Thus $C_E(A)$ is a soluble group. Its preimage in $G_0$ is an extension by $R$, so it too is soluble and has finite index in $G$. Virtually soluble groups are closed under group extensions.

For the final example take the restricted direct sum of groupswhere $A_5$ is the nonabelian simple group of even permutations on five letters. Every finite collection of elements lies in a product of finitely many finite factors, so $D$ is a locally finite group and hence a torsion group. It cannot contain a nonabelian free group, which is a torsion-free group.

To show that $D$ is not a virtually soluble group, let $H$ be any finite-index subgroup and let $N\le H$ be the kernel of the finite coset action. Each coordinate $A_5$ maps either injectively or trivially into the finite quotient $D/N$, by Simplicity of the alternating group A5. The nontrivial images of distinct factors commute, and each has trivial centre, so any $k$ of them generate a direct product of groups of order $60^k$. Only finitely many such images can occur in a finite quotient. Therefore $N$, and hence $H$, contains a whole coordinate copy of $A_5$, which is not soluble: its nontrivial commutator subgroup is normal and therefore equals $A_5$. No finite-index subgroup of $D$ is soluble.

Conversely, assume $H$ is a normal subgroup and the quotient group $G/H$ is an abelian group. Commuting the two cosets $aH$ and $bH$ givesso every group commutator $a^{-1}{b}^{-1}ab$ belongs to $H$. This proves the converse and completes the equivalence.

For the dihedral group of order ten, choose a rotation $r$ and a reflection $s$ with $r^5=s^2=1$ and $srs=r^{-1}$. Its rotation subgroup $⟨r⟩$ is a normal subgroup, and the quotient by it has order two, hence is an abelian group. Consequently the commutator subgroup is contained in $⟨r⟩$. On the other hand,Since $r^{-2}$ generates the order-five cyclic group $⟨r⟩$, the commutator subgroup is exactly $⟨r⟩$. Every abelian quotient must kill this subgroup. There are only two subgroups containing it, since the remaining quotient has prime order. The complete list, including the trivial quotient, isThis is the order-ten instance of the abelianization of an odd dihedral group.

Suppose every group commutator belongs to $H$. For $h\in H$ and $g\in G$, the particular group commutator $h^{-1}{g}^{-1}hg$ belongs to $H$. Multiplying on the left by $h$ gives $g^{-1}hg\in H$. Thus $g^{-1}Hg\subseteq H$ for every $g$; applying the same inclusion to $g^{-1}$ gives the reverse inclusion. Therefore $H$ is a normal subgroup.

We can now form the quotient group. Its group commutators areA group commutator is the identity element exactly when the two elements commute. Hence $G/H$ is an abelian group. Equivalently, the hypothesis says that $H$ contains the commutator subgroup, so the quotient factors through the abelianization.

Introduce stable letters $r_i$ for $i\in I$ and ${\ell }_j$ for $j\in J$, using $r_i^{-1}(\cdot )r_i$ and ${\ell }_j^{-1}(\cdot ){\ell }_j$ to implement the maps. Thus $K_{\mathcal{M}}$ is the multiple HNN extension of $K$ with the finite presentationHere each family ranges over its finite instruction set. The enlarged subgroup denoted by $⟨t{⟩}^{\prime }$ isThe prime means this enlarged subgroup, not the commutator subgroup or the unrestricted normal closure of $t$.

The key relationship isHere is a justification that also identifies the relevant intersection. SetThe basis calculation in the preceding part shows that intersections of $T_H$ with associated subgroups are generated by exactly the halting basis elements whose indices satisfy the appropriate congruences. Since the machine is deterministic and $(0,0)$ is terminal, halting at $(0,0)$ is equivalent for the two ends of every instruction edge. Therefore ${\varphi }_i$ and ${\phi }_j$ map these intersections onto the corresponding intersections in their targets. The same assertion holds on the integer lattices after declaring pairs with a negative coordinate nonhalting: the bounds on $a,b,c$ imply that each coordinate's sign is preserved by a transition and its inverse.

Apply Britton's lemma to a word in $B_H$ which represents an element of $K$. If it has stable letters, it must have a pinch. The base coefficient of that pinch lies in $T_H$ and an associated subgroup, and the intersection property just proved makes its transported image lie in $T_H$ again. Successive pinch removals therefore keep all base coefficients in $T_H$, and eventually remove every stable letter. It follows thatFinally, $t=t(0,0)$ belongs to $T_H$, so $L\subseteq B_H$. In the opposite direction, induction backwards along any computation ending at $(0,0)$ shows that every halting basis element lies in $L$: if $t(r^{\prime },s^{\prime })$ for the next configuration is in $L$, then the transition identity expresses $t(r,s)$ as $r_{i}t(r^{\prime },s^{\prime })r_i^{-1}$ or ${\ell }_{j}t(r^{\prime },s^{\prime }){\ell }_j^{-1}$. Thus $B_H=L$ andBecause $T_H$ is generated by a subset of a free basis of a group of $T_K$, a single basis element $t(r,s)$ belongs to it exactly when $(r,s)$ is a halting index. This proves the boxed relationship.

The commutator subgroup, or derived subgroup, isA linear character is afforded by a one-dimensional representation $\rho :G\to {\mathbb{C}}^{\times }$. Since ${\mathbb{C}}^{\times }$ is abelian, every commutator lies in $\ker \rho$, so

Let $\pi :G\to G/G^{\prime }$ be the quotient map. Every irreducible representation of a finite abelian group is one-dimensional, so each irreducible character $\psi$ of the abelianization $G/G^{\prime }$ inflates to the linear character $\psi \circ \pi$ of $G$. Conversely, the containment $G^{\prime }\le \ker \chi$ makes every linear character of $G$ factor uniquely through $G/G^{\prime }$. Thus inflation of a group representation gives a bijection

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.

For $n\ge 2$, the commutator subgroup of $S_n$ is $A_n$. The group $A_5$ is nonabelian and simple, so its commutator subgroup is a nontrivial normal subgroup and must equal $A_5$ itself. Hence the derived series of $S_5$ isand never reaches the identity. Therefore

The commutator subgroup $[G,G]$ is characteristic: every automorphism sends commutators to commutators and therefore preserves the subgroup they generate. Thus an automorphism $A$ inducesThis is independent of the coset representative $g$. If $A^{\prime }=c_h\circ A$ differs from $A$ by an inner automorphism, thenbecause the abelianization is abelian. Hence $q([A])=\overline{A}$ is independent of the representative of the outer class. Finally $\overline{AB}=\overline{A}\overline{B}$, sois a well-defined group homomorphism.

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero quadratic residue modulo $11$. The square subgroup isMatrix multiplication becomesThus $H$ is the affine semidirect product of cyclic groups of orders eleven and fiveThere are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since

LetConjugation by the complement givesHence the ten nonidentity translations split into the two orbitseach of size five. For $r\ne 1$,Since $1-r\ne 0$, varying $t$ gives every $b\in {\mathbb{F}}_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, withthe seven conjugacy classes areof sizes $1,5,5,11,11,11,11$, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in ${\mathbb{F}}_{11}$. Hence the commutator subgroup is $N$, and the abelianization isPut $\zeta =e^{2\pi i/5}$. The one-dimensional characters factor through the abelianization, giving five characters

For the remaining characters, put $\omega =e^{2\pi i/11}$ and define a character of $N$ byThe complement has two free orbits on the nontrivial ${\theta }_m$, indexed by $m\in Q$ and $m\in 2Q$. By induction from an abelian normal subgroup with a free character orbit, the charactersare irreducible of degree five and vanish outside $N$.

SetThe quadratic periods modulo eleven give the nonsquare sum $\overline{\eta }=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete character table isFinally,and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

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.