= Solution
A <soluble group>, also called a <solvable group>, has a terminating <derived series>:
$$
G^{(0)}=G,\qquad G^{(j+1)}=[G^{(j)},G^{(j)}],\qquad G^{(r)}=1\text{ for some }r\geq0.
$$
For a <subgroup> $S\leq G$, induction gives $S^{(j)}\leq G^{(j)}$, 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> $1\to K\to E\to Q\to1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\leq K$ and $E^{(s+r)}=1$. \b[<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 $S\leq R$. Thus $R$ is the unique largest soluble <normal subgroup> of $K$, making it a <characteristic subgroup>, and it has finite index.
Now suppose $1\to K\to G\overset{\pi}{\to}Q\to1$ 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 $E\to\operatorname{Aut}(A)$ 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 $G$. \b[<Virtually soluble groups> are closed under <group extensions>.]
For the final example take the <restricted direct sum of groups>
$$
\boxed{D=\bigoplus_{j\geq1}A_5,}
$$
where $A_5$ 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 $N\leq H$ 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 $A_5$. \b[No finite-index subgroup of $D$ is soluble.]
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