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Every finite group whose order is divisible by at most two distinct primes is a solvable group. A nonabelian simple group must therefore have at least three distinct prime divisors.

Every finite solvable group is isomorphic to a subgroup of a finite monomial group. This is an embedding assertion and does not assert that every solvable group is monomial, nor that the embedding is normal.

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.

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

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

A nontrivial minimal normal subgroup $N$ of a finite solvable group is abelian: its commutator subgroup is characteristic, and minimality plus solvability excludes $N^{\prime }=N$. A nontrivial Sylow subgroup of this abelian group is characteristic, so minimality makes $N$ a p-group. The characteristic subgroup of pth powers is proper, hence trivial by minimality. Therefore $N$ is an elementary abelian group.

Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a soluble group if this derived series reaches the identity after finitely many steps. Its derived length is the least $d$ with $G^{(d)}=1$.

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. The group $A_4$ is nonabelian, so its derived subgroup is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived subgroup is a normal subgroup, containing any one forces it to contain all three. Thus $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, every commutator in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a subgroup of a solvable group is solvable, with no larger derived length.

The sign map gives $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.

A finite monomial group, also called an M-group, is a finite group for which every irreducible character is a monomial character: for every $\chi \in \mathrm{Irr}(G)$ there are a subgroup $H\le G$ and a linear character $\lambda$ of $H$ such that $\chi ={\mathrm{Ind}}_H^G\lambda$. The inducing subgroup and linear character may depend on $\chi$.

Taketa's theorem states that every finite monomial group is solvable. Here being a solvable group means that the iterated derived subgroups eventually reach the identity.

Dade's embedding theorem for monomial groups states that every finite solvable group embeds as a subgroup of a finite monomial group. It does not say that every solvable group is itself monomial, nor that the embedding can always be taken normal. Together the results identify solvability as a necessary property of monomial groups without making monomiality closed under taking arbitrary subgroups.

In character theory, a p-elementary group is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An elementary group is p-elementary for at least one prime $p$; it need not be an elementary abelian group.

A generalised character means an integer linear combination of irreducible complex characters. Brauer's characterisation of characters states that a complex class function $f$ on a finite group $G$ is a generalised character if and only if $f_E$ is a generalised character for every elementary subgroup $E\le G$. Equivalently,An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.

For the extension construction, set $d=\theta (1)$. The determinant character $\det \theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a linear character. For any $g\in G$, the subgroup $J_g=N⟨g⟩$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and ${\mu }_{J_g}$ extends its determinant. The permitted coprime-degree determinant extension theorem therefore gives a unique character ${\chi }_g\in \mathrm{Irr}(J_g)$ withDefine a function on $G$ by $f(g)={\chi }_g(g)$. We next verify that these individually defined values form a compatible class function.

If $x\in G$, conjugation carries $J_g$ to $J_{xg{x}^{-1}}$. Transporting ${\chi }_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a class function. Uniqueness identifies the transported character with ${\chi }_{xg{x}^{-1}}$. Consequently $f(xg{x}^{-1})=f(g)$.

Now take an elementary subgroup $E\le G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of solvable groups is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives ${\chi }_J\in \mathrm{Irr}(J)$ extending $\theta$ with determinant ${\mu }_J$.

For $g\in E$, we have $J_g\le J$. The restriction $({\chi }_J{)}_{J_g}$ is irreducible: an invariant subspace for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is ${\mu }_{J_g}$. By uniqueness on $J_g$, it equals ${\chi }_g$. Thereforean ordinary character and hence a generalised character of $E$. Brauer's characterisation of characters now makes $f$ a generalised character of $G$. If $n\in N$, then $J_n=N$ and ${\chi }_n=\theta$, so $f(n)=\theta (n)$. We have proved, by gluing determinant-normalized character extensions,Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.

For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_{\pi }\times C_{{\pi }^{\prime }}$, where each factor has the indicated prime divisors in its order. If $p\in \pi$, put $E_{\pi }=P\times C_{\pi }$ and $E_{{\pi }^{\prime }}=C_{{\pi }^{\prime }}$. If $p\notin \pi$, put $E_{\pi }=C_{\pi }$ and $E_{{\pi }^{\prime }}=P\times C_{{\pi }^{\prime }}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the prime-set decomposition of elementary groupsHere a pi-group has order divisible only by primes in $\pi$, and a pi-element has such an element order; the identity qualifies for both complementary prime sets.

Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of conjugacy classes and are disjoint, because membership in both would force every prime divisor of the element order into $\pi \cap {\pi }^{\prime }=\varnothing$, hence force order $1$. Let $E\le K$ be elementary and use $E=E_{\pi }\times E_{{\pi }^{\prime }}$. If both factors were nontrivial, choose $a\ne 1$ in the first and $b\ne 1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and ${\pi }^{\prime }$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a pi-group or entirely a complementary-prime group.

Write $m=|K{|}_{\pi }$ and $r=|K{|}_{{\pi }^{\prime }}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd (m,r)=1$. The Chinese remainder theorem supplies an integer $D$ satisfyingDefine the class functionFor an elementary $\pi$-subgroup $E$, its order divides $m$. Let ${\rho }_E$ be the character of its regular representation, equal to $|E|$ at $1$ and zero elsewhere. Thenwith integral coefficient because $D\equiv 1(\mathrm{mod}|E|)$. For an elementary ${\pi }^{\prime }$-subgroup, its order divides $r$, andagain has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a generalised character. A second application of Brauer's characterisation of characters provesThe freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi =\varnothing$, ${\pi }^{\prime }=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.

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.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.

A group is solvable when its derived serieseventually reaches the trivial group; equivalently, it has a subnormal series with abelian quotients.

The symmetric group $S_4$ is solvable becausehas abelian quotients $V_4$, $C_3$, and $C_2$, where $V_4$ is the Klein four-group. Thus

Let $x_1,\dots ,x_5$ be algebraically independent indeterminates and set $a_i=e_i(x_1,\dots ,x_5)$, the elementary symmetric polynomials. ThenThe Fundamental theorem of symmetric polynomials identifies the fixed field of the natural $S_5$ action on $\mathbb{Q}(x_1,\dots ,x_5)$ with $\mathbb{Q}(a_1,\dots ,a_5)=K$. Every permutation of the roots therefore gives a $K$-automorphism, and the Galois group of a polynomial isA characteristic-zero polynomial is solvable by radicals exactly when its Galois group is a solvable group. Part (c) shows that $S_5$ is not solvable, so

A finite solvable group is a group admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chainhas quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every subgroup of a solvable group is solvable, because intersecting the chain with the subgroup gives factors isomorphic to subgroups of the original abelian factors. A Galois group of a polynomial of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the polynomial is reducible or has repeated roots.

For the substitution, the characteristic-zero solvability by radicals theorem says that a polynomial has solvable Galois group if and only if its splitting field is contained in a finite tower formed by adjoining successive roots of elements of the preceding field. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th root of unity and, for every nonzero $\alpha$, an $m$th root ${\beta }_{\alpha }$. These are radical adjunctions, and all roots of $x^m-\alpha$ are ${\beta }_{\alpha }{\zeta }_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the splitting field of $f(x^m)$. The theorem implies

A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

A simple group is a nontrivial group whose only normal subgroups are itself and the trivial subgroup.

A composition series is a chain $G=G_0▹G_1▹\cdots ▹G_r=1$ in which each $G_{i+1}$ is a normal subgroup of $G_i$ and each quotient group $G_i/G_{i+1}$ is simple and nontrivial. Its length is $r$, and these quotients are its composition factors. The individual terms need not all be normal in $G$.

A solvable group is a group whose derived series reaches the trivial group: starting with $G^{(0)}=G$, define $G^{(i+1)}=[G^{(i)},G^{(i)}]$. For finite groups this is equivalent to having a composition series whose factors are all cyclic groups of prime order.

Factor the group order as $588=2^2\cdot 3\cdot 7^2$. The Sylow theorems say that the number of Sylow subgroups for the prime seven divides $12$ and is congruent to one modulo seven. Among $1,2,3,4,6,12$, only one qualifies. Thus there is a unique normal subgroup $N$ of order $49$.

Every group of prime-square order is Abelian. Indeed its center of a group is nontrivial, and a cyclic quotient by the center of a group forces the whole group to be abelian. Consequently $N$ is either $C_{49}$ or $C_7\times C_7$, and in either case it has two composition factors, both $C_7$.

It remains to consider $Q=H/N$, of order twelve. Its number of Sylow subgroups for the prime three is either one or four:

Lift the composition series of $Q$ to $H$ using the subgroup correspondence theorem, and join it to a composition series inside $N$. This givesas the multiset of composition factors. The Jordan–Hölder theorem makes this independent of the chosen series. In particular every group of order $588$ is a solvable group, with composition length five; that is a statement about composition length, not necessarily its derived length.

Write $|G|=pq$ with $p<q$. The Sylow theorems give $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. Since $p<q$, necessarily $n_q=1$. Its unique Sylow subgroup $N$ is a normal subgroup and has prime order $q$, soA Sylow subgroup $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. ThusThe semidirect product action is a group homomorphism $C_p\to \mathrm{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the Chinese remainder theorem; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a metacyclic group and hence a solvable group.

A finite separable extension is solvable when its normal closure has a solvable group as Galois group. In characteristic zero it is contained in a radical tower after adjoining suitable roots of unity; containment need not mean it is itself a tower of radical adjunctions.

Every finite monomial group is a solvable group. Thus expressing every irreducible character as induction from a linear character imposes a group-theoretic solvability condition.