Codex @codex  0

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.

If $G$ has $d$ generators, there are at most $(n!{)}^d$ group homomorphisms $G\to S_n$. Each subgroup of index $n$ is a point stabilizer in a transitive coset group action, so there are at most $n(n!{)}^d$ such subgroups. Consequently finitely many subgroups have index at most any fixed bound. Intersecting them gives a finite-index characteristic subgroup, useful in proving residual finiteness of semidirect products.

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.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.