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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.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

A permutation representation is a group homomorphism $\rho :G\to \mathrm{Sym}(\Omega )$, equivalently a group action on a set $\Omega$. The left regular action takes $\Omega =G$ and $\rho (g)(h)=gh$. Each such map is a bijection, and $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$. If $\rho (g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\mathrm{Sym}(G)$ with the symmetric group $S_n$, proving Cayley theorem:For a nonabelian simple group, compose this faithful representation with the sign of a permutation homomorphism $S_n\to \{1,-1\}$. Its kernel is a normal subgroup of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and the same representation embeds $G$ in the alternating group $A_n$.

Finally let $S_3$ act on the two cosets of its normal subgroup $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with kernel $A_3$; explicitly, send even permutations to the identity and odd permutations to $(12)$.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.

Burnside's theorem states that every finite group of order $p^a{q}^b$ is solvable, where $p,q$ are primes and $a,b$ are nonnegative integers. In particular, a nonabelian simple group cannot have order divisible by only one or two distinct primes.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Multiplicativity of the determinant gives $\delta (gh)=\delta (g)\delta (h)$ and $\delta (1)=1$, so $\delta$ is a linear character. Its finite image is a cyclic group of roots of unity. If it contains $-1$, its order is $2m$ and the character of a representation ${\delta }^m$ maps onto $\{\pm 1\}$. The kernel of a group homomorphism is the preimage of its identity, so $\ker ({\delta }^m)$ is a normal subgroup of index $2$. The subgroup $\ker \delta$ itself need not have index $2$.

By Cauchy theorem for groups, a group $E$ of order $2k$ has an involution $g$. In the regular representation, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its determinant is $(-1{)}^k=-1$ when $k$ is odd. The preceding argument gives a normal subgroup of index two.

For a nonabelian simple group of order below $80$, Burnside's theorem requires at least three distinct prime factors. An odd order with three factors is at least $3\cdot 5\cdot 7=105$. An even order with exactly one factor of $2$ has a normal subgroup of index $2$. Thus its order must be divisible by $4$ and by at least two odd primes. Below $80$ the only possibility is $4\cdot 3\cdot 5=60$, since the next is $4\cdot 3\cdot 7=84$. Hence $\boxed{|H|=60}$.

A nonabelian simple group is a perfect group, so every one-dimensional group representation, a homomorphism to the abelian group $k^{\times }$, is trivial. Thus a nontrivial irreducible Brauer character cannot have degree $1$.

Suppose instead that its representation has dimension $2$, working over a splitting extension if necessary. Its determinant is a one-dimensional representation and is therefore trivial. Its kernel is a normal subgroup, and the representation is nontrivial, so simplicity makes it faithful. Consequently $G$ embeds in ${\mathrm{SL}}_2(k)$.

We use the Feit–Thompson theorem: every finite group of odd order is solvable. Therefore this nonabelian simple group has even order, and Cauchy's theorem for finite groups supplies an involution $t$. In odd characteristic, its representing matrix $A$ satisfies $A^2=I$ and is diagonalizable with eigenvalues in $\{1,-1\}$. Since $\det A=1$, it is $I$ or $-I$. Faithfulness excludes $I$, so $t$ acts as the scalar matrix $-I$. It commutes with the whole image; faithfulness then makes $t$ central in $G$, contradicting nonabelian simplicity. Hence $\boxed{\chi (1)>2}$.

The odd order theorem is the deep standard group-theoretic input in this proof; its use is explicit rather than hidden in an unsupported assertion that $G$ has an involution.

The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case