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Work in the finite-group setting. If $M$ is maximal in a nonabelian simple $G$, then $M\ne 1$: otherwise every nonidentity element would generate $G$, forcing a cyclic group of prime order. Choose a minimal normal subgroup $N\ne 1$ of $M$. Every characteristic subgroup of $N$ is normal in $M$, so minimality makes $N$ a characteristically simple group. Its normalizer contains $M$. By maximality it is $M$ or $G$, and the latter would make $N$ a nontrivial normal subgroup of the simple group $G$, impossible because $N\le M<G$. This proves the maximal subgroup normalizer criterion:

For $A_5$, stabilizer subgroups have order 12. Stabilizer subgroups of a two-element subset have order $|(S_2\times S_3)\cap A_5|=6$, and are isomorphic to $S_3$; for example $⟨(123),(12)(45)⟩$. The normalizer of a Sylow 5-subgroup has order 10: its normalizer in $S_5$ has order $5\cdot 4$, and exactly half its permutations are even, since a multiplier of order four on ${\mathbb{F}}_5$ is an odd four-cycle. This gives a dihedral group of order 10.

To prove completeness, an intransitive subgroup either fixes a point, or its orbits on five letters have sizes two and three. It therefore lies in one of the first two types. A transitive proper subgroup $H$ has order divisible by 5. Its number of Sylow 5-subgroups is one or six, since $A_5$ has six and the count is $1$ modulo 5. A count of six would make $30\mid |H|$, forcing an index-two subgroup and contradicting Simplicity of the alternating group A5. Therefore $H$ normalizes its unique Sylow 5-subgroup and lies in the order-ten normalizer.

Stabilizer subgroups are maximal because their index is prime. The order-ten group is transitive and cannot have a larger proper transitive overgroup by the preceding argument. The order-six group has no fixed point, cannot lie in a stabilizer subgroup, and cannot lie in the order-ten type because 3 does not divide 10. Thus all three types are maximal. $A_5$ is transitive on points and on unordered pairs, while Sylow conjugacy handles the normalizers. Consequently

Now let $G=G{L}_3(2)$. Choosing independent columns gives $|G|=7\cdot 6\cdot 4=168$. Its simplicity follows from the elementary projective-transvection proof in part 4(i), which does not use this subgroup classification.

A stabilizer subgroup and a plane stabilizer subgroup each have index seven and order 24. To construct the remaining type, identify ${\mathbb{F}}_2^3$ with ${\mathbb{F}}_8$ and take the Singer cycle $P$ of multiplications by nonzero field elements, of order seven. Any linear map commuting with a primitive multiplication commutes with all of ${\mathbb{F}}_8$, so its centralizer is precisely $P$. A normalizer sends a primitive element $a$ to another root of its minimum polynomial, namely $a,a^2,a^4$. The Frobenius map $x\mapsto x^2$ realizes those three choices. Hence

If $7\mid |H|$ for a proper $H<G$, its Sylow count is one or eight. Eight would make $56\mid |H|$, hence $|H|=56$. Its index-three coset action would inject the simple group $G$ into $S_3$: the kernel is normal and cannot be all of $G$. This is impossible. Thus $H$ normalizes a unique Sylow 7-subgroup and lies in the order-21 normalizer.

If $7\nmid |H|$, its order divides 24. In its action on seven nonzero vectors, an odd orbit has size one or three. A one-point orbit gives a fixed point. If a three-point orbit is collinear, it is the set of nonzero vectors of an invariant plane. If its three vectors are independent, their nonzero sum is fixed by $H$. Thus $H$ lies in a point or plane stabilizer subgroup.

The index-seven stabilizer subgroups are maximal, and any proper overgroup of the order-21 normalizer would again have a unique Sylow 7-subgroup and normalize that same subgroup, so the normalizer is maximal too. The point and plane stabilizer subgroups are not conjugate: a stabilizer subgroup fixes a point, whereas a plane stabilizer subgroup is transitive on its three internal points and its four external points, so fixes no point. Each type is one conjugacy class. We have proved maximal subgroups of GL3 over F2:

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.

Use the field $F={\mathbb{F}}_2[\zeta ]/({\zeta }^3+\zeta +1)$. The defining polynomial has no root in ${\mathbb{F}}_2$, hence is irreducible. Label its elements bySince $F^{\ast }$ has prime order seven, $\zeta$ has order seven. Multiplication by $\zeta$ is exactly $b$. The identities ${\zeta }^3=\zeta +1$, ${\zeta }^6={\zeta }^2+1$ and ${\zeta }^5={\zeta }^4+1$ show that translation by one is exactly $a$.

Conjugating $a$ by powers of $b$ gives the translations $T_{{\zeta }^i}:z\mapsto z+{\zeta }^i$. The translations by $1,\zeta ,{\zeta }^2$ generate all eight translations. ConsequentlyFor distinct $u,v$ and distinct target points $u^{\prime },v^{\prime }$, the unique affine map has $s=(v^{\prime }-u^{\prime })/(v-u)$ and $t=u^{\prime }-su$. Thus the action is sharply two-transitive.

Its regular characteristic subgroup isTranslations act regularly, $K$ is normal, and its order eight makes it the unique Sylow two-subgroup, hence characteristic. This identifies the abstract subgroup and explicit generators in the original permutation notation.

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.

The largest normal p-subgroup of a finite group. Products of normal p-subgroups are again normal p-subgroups, since $|AB|=|A||B|/|A\cap B|$. Thus the subgroup generated by all such subgroups is itself a p-subgroup. It is a characteristic subgroup.