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A Hall subgroup for a set of primes $\pi$ is a subgroup $H\le G$ such that every prime divisor of $|H|$ belongs to $\pi$, while no prime divisor of $[G:H]$ belongs to $\pi$. Equivalently $|H|=|G{|}_{\pi }$, the product of the full prime powers in $|G|$ belonging to $\pi$.

The Hall conjugacy and embedding in finite soluble groups theorem says that, for every $\pi$, a finite soluble group has a Hall $\pi$-subgroup, any two such subgroups are conjugate, and every $\pi$-subgroup is contained in one. We prove all three assertions together by induction, including the needed complement argument rather than assuming it.

First establish coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary Abelian $r$-group and $D=E/V$ has order $m$ prime to $r$. Write $V$ additively. Conjugation defines a well-defined action of $D$ on $V$ because $V$ is Abelian. Choose a section $s:D\to E$, with $s(1)=1$, and let its extension cocycle be defined by $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in D$. With $S(x)={\sum }_{z}f(x,z)$ this becomesMultiplication by $m$ is invertible on $V$, so setting $b(x)=m^{-1}S(x)$ gives $f(x,y)=b(x)+xb(y)-b(xy)$. The new section $s^{\prime }(x)=(-b(x))s(x)$ has zero cocycle and is a homomorphism. Its image is a complement $K$ to $V$, proving $E=V⋊K$.

Every other complement has the form $\{h(x)s^{\prime }(x):x\in D\}$, with $h(xy)=h(x)+xh(y)$. Summing this identity over $y$ and putting $T={\sum }_{y}h(y)$ gives $mh(x)=T-xT$. Hence $h(x)=v-xv$, where $v=m^{-1}T$, and the other complement is $vK{v}^{-1}$. The same argument applies on any subgroup $D_0\le D$: a subgroup intersecting $V$ trivially is a graph of such a one-cocycle over its image $D_0$, so conjugation by an element of $V$ carries it into $K$. This proves the complement-conjugacy and embedding assertions needed below.

Now let $G\ne 1$ be finite soluble, and choose a minimal normal subgroup $V$. It is elementary Abelian of some prime characteristic $r$, by the minimal-normal argument in the series discussion. Induction gives Hall subgroups and the embedding and conjugacy assertions in $G/V$. Let $\overline{H}$ be a Hall $\pi$-subgroup there and $M$ its full preimage.

If $r\in \pi$, then $M$ itself is a Hall $\pi$-subgroup of $G$. Every Hall $\pi$-subgroup $H$ contains $V$, since $HV$ is a $\pi$-subgroup and $H$ already has the largest possible $\pi$-order. Thus $H$ is the full preimage of its image in $G/V$, and quotient conjugacy lifts to conjugacy in $G$. Given any $\pi$-subgroup $U$, induction puts its quotient image into some $\overline{H}$, whose preimage $M$ contains $U$.

If $r\notin \pi$, then $M/V=\overline{H}$ has order prime to $r$. The complement construction produces $K\le M$ of order $|\overline{H}|=|G{|}_{\pi }$, hence a Hall subgroup of $G$. Any Hall subgroup has trivial intersection with $V$ and a Hall image in $G/V$. Conjugating two Hall subgroups first aligns these images, after which both are complements to $V$ in the same $M$; the complement-conjugacy argument then conjugates them by an element of $V$. For an arbitrary $\pi$-subgroup $U$, quotient induction puts its image into some $\overline{H}$. Then $U\subseteq M$ and $U\cap V=1$, so the embedding argument conjugates $U$ into the chosen complement $K$. Equivalently $U$ lies in a conjugate of $K$. This completes the induction and all three parts of the Hall theorem for soluble groups.

For the specific general linear group over a finite field, counting ordered bases givesThere are seven lines and seven planes in ${\mathbb{F}}_2^3$. The group is transitive on each family, so line and plane stabilizers both have order $24$ and index seven. To show that these exhaust the index-seven subgroups, let $H$ have order $24$. It contains a Sylow $2$-subgroup of order eight, which can be conjugated toThe group orbits of $U$ on the seven nonzero vectors areof sizes $1,2,4$. An $H$-orbit is a union of these. The group $H$ cannot be transitive on seven points, since seven does not divide $24$. Therefore it has an orbit of size at most three. A size-one orbit fixes $e_1$. A size-two orbit must be $\{e_2,e_1+e_2\}$, whose vector sum $e_1$ is consequently fixed. A size-three orbit must be the union of the first two $U$-orbits, the nonzero part of the plane $⟨e_1,e_2⟩$, so that plane is invariant. In every case $H$ lies in a line or plane stabilizer, and equality follows from their equal orders.

The two stabilizer families are not conjugate. A line stabilizer fixes its nonzero vector and is transitive on the other six nonzero vectors: extend that fixed vector and either chosen other vector to a basis, and map one such basis to the other. Thus it has no invariant three-element subset of nonzero vectors and fixes no plane. In contrast a plane stabilizer fixes its defining plane. Conjugation sends line stabilizers to line stabilizers, so cannot send one to a plane stabilizer. We obtainThey are Hall subgroups for $\pi =\{2,3\}$ but are not conjugate. This explicitly shows why the solubility hypothesis of the Hall theorem for soluble groups matters.

Finally, an index-three subgroup would give a nontrivial transitive coset homomorphism $G{L}_3(2)\to S_3$. Its kernel is normal; the permitted simplicity of $G{L}_3(2)$ forces that kernel to be trivial, because the action is nontrivial. But an injective map from a group of order $168$ into one of order six is impossible. Hence $G{L}_3(2)$ has no subgroup of index three.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

Assume first that the topological group is Hausdorff. This separation condition is necessary: an indiscrete group with two elements has the one-element neighbourhood basis consisting of the whole group, but its topology cannot come from a metric.

Translations are homeomorphisms, since multiplication by a fixed element is continuous and its inverse is multiplication by the inverse element. Inversion is likewise a homeomorphism. If $W$ is an neighbourhood of the identity element, continuity of the triple product at $(e,e,e)$ supplies neighbourhoods of the identity element $A,B,C$ with $ABC\subset W$. Their intersection with their inverses contains a symmetric open neighbourhood of the identity element $V$, and $V^3\subset W$. This proves the neighbourhood shrinking property needed below rather than assuming it.

Let $(B_n)$ be a countable neighbourhood basis at the identity element, choosing open representatives inside its members. Starting with $U_0=G$, recursively choose symmetric open neighbourhoods of the identity element such thatThey are still a basis. Their intersection is $\{e\}$: for $x\ne e$, the Hausdorff condition gives an neighbourhood of the identity element not containing $x$, and some $U_n$ lies inside it.

Here is the dyadic product estimate for group neighbourhoods. If $g_j\in U_{n_j}$ and ${\sum }_j{2}^{-n_j}\le 2^{-n}$, then the ordered product lies in $U_{n-1}$, for $n\ge 1$. Prove this by induction on the number of factors. A single factor has $n_j\ge n$ and hence belongs to $U_n\subset U_{n-1}$. For a longer word, choose the factor crossing the half-total-cost point. The words preceding and following it each have cost at most half the total, hence at most $2^{-n-1}$. The induction hypothesis puts both in $U_n$, and the middle factor also lies in $U_n$. Their threefold product is in $U_{n-1}$. Empty side words cause no problem because $e\in U_n$.

Give an edge from $x_{j-1}$ to $x_j$ cost $2^{-n_j}$ whenever $x_{j-1}^{-1}{x}_j\in U_{n_j}$, and defineChains exist because $U_0=G$. Reversing a chain proves symmetry, concatenating chains proves the triangle inequality, and translating every vertex on the left leaves all edge differences unchanged. Thus $d$ is left invariant. The product estimate also givesThe first implication follows by choosing a chain with total cost below the bound and applying the estimate with index $n+1$. If $d(x,y)=0$, the first implication puts $x^{-1}y$ in every $U_n$, so $x=y$. Hence $d$ is a genuine left-invariant group metric. These two inclusions show that its balls and the original neighbourhoods of the identity element refine one another; translations then show equality of the two topologies everywhere. More explicitly, every metric ball contains a sufficiently small translated $U_n$, and every open neighbourhood contains a sufficiently small metric ball. This proves the sufficient direction of the Birkhoff-Kakutani theorem.

Conversely, if a compatible metric exists, the balls $B_d(e,1/n)$ form a countable neighbourhood basis at the identity element. Invariance is not even needed for this direction. Thus, with the necessary Hausdorff convention, a compatible left-invariant metric exists exactly when the identity element has a countable neighbourhood basis.

For the requested counterexample, take the orientation-preserving affine group of the real line, represented by $(a,b)$ with $a>0$ and product $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$. Its topology is metrizable, for instance by the ordinary Euclidean distance in coordinates $(\log a,b)$, and the product and inverse are continuous. Let $D_a=(a,0)$ and $T_b=(1,b)$. Direct multiplication gives $D_a{T}_b{D}_a^{-1}=T_{ab}$. A compatible bi-invariant group metric would make every conjugation an isometry fixing $e$. ThereforeBut $T_{1/n}\to e$ in the group's topology, a contradiction. This is the affine conjugation obstruction to a bi-invariant group metric: the group is metrizable, and even admits a compatible left-invariant metric, but admits no compatible metric invariant on both sides.

Let $G$ be locally compact, Abelian and Hausdorff, and fix a Haar measure. Write $T_{x}f(y)=f(x^{-1}y)$ and use convolutionThe L1 group algebra is a commutative Banach algebra. Here a group character means a continuous homomorphism $\chi :G\to \mathbb{T}$, not the trace of an arbitrary group representation. A Gelfand character is a nonzero multiplicative complex-linear functional on the convolution algebra. We use the conventionWith the alternative standard Fourier convention involving $\overline{\chi }$, one simply conjugates every group character in this identification.

For such $\chi$, the functional is bounded by $\Vert f{\Vert }_1$, and Fubini's theorem, the substitution $y=xz$, and the homomorphism law giveIt is nonzero: if $u\ge 0$ is nonzero in $C_c(G)$, then $f=\overline{\chi }u$ has ${\Phi }_{\chi }(f)=\int u>0$.

Conversely let $\Phi$ be an algebra character. It is automatically bounded, with $|\Phi (f)|\le \Vert f{\Vert }_1$. Indeed, extend it to the unitization by $\tilde{\Phi }(f+\lambda 1)=\Phi (f)+\lambda$. If $|\Phi (f)|>\Vert f{\Vert }_1$, the Neumann series makes $f-\Phi (f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.

Choose $g$ with $\Phi (g)\ne 0$ and put $\chi (x)=\Phi (T_{x}g)/\Phi (g)$. Commutativity of the group gives $(T_{x}f)\ast g=f\ast (T_{x}g)$; applying $\Phi$ yieldsThus the ratio is independent of the chosen $g$. Applying the relation successively to $T_x{T}_{y}g=T_{xy}g$ gives $\chi (xy)=\chi (x)\chi (y)$, and $\chi (e)=1$. In particular it never vanishes.

The translation map $x\mapsto T_{x}g$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one compact set; continuity on that set and a finite cover give uniform convergence of the translated values, hence $L^1$ convergence because the compact set has finite Haar measure. For general $g$, approximate by $C_c(G)$ and use $\Vert T_{x}g{\Vert }_1=\Vert g{\Vert }_1$. Therefore $\chi$ is continuous. For all integers $n$,Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi (x)|=1$.

Finally, the $L^1$-valued convolution formula $f\ast g=\int f(x)T_{x}gdx$ and boundedness of $\Phi$ giveCancel $\Phi (g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small neighbourhood detects that difference. Thus the correspondence is bijective.

It also identifies the character space of an Abelian L1 group algebra topologically with the Pontryagin dual group. For a net converging uniformly on compact subsets in $\hat{G}$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common norm bound one. Conversely suppose the functionals converge in the Gelfand topology. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_{x}g:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded linear functionals is uniform on this compact set: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives uniform convergence of the group characters on $K$. This proves equality with the compact-open topology, without assuming first countability of $G$.

For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\hat{G})$. The set of algebra characters together with the zero functional is weak-star compact by Banach-Alaoglu theorem and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff character space; each evaluation $\Phi \mapsto \Phi (f)$ vanishes at infinity, since its modulus-at-least-$ϵ$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under conjugation because $f^{\ast }(x)=\overline{f(x^{-1})}$ satisfies ${\Phi }_{\chi }(f^{\ast })=\overline{{\Phi }_{\chi }(f)}$. The locally compact Stone-Weierstrass theorem therefore gives the claimed density.

For a subgroup $H$ of a finite group $G$, Lagrange's theorem states $|G|=[G:H]|H|$. Thus a subgroup's order, and in particular the order of a group element, divides $|G|$.

For $|G|=10$, if every element had order one or two, the order of a finite group of exponent two would make $|G|$ a power of two, a contradiction. Hence an element has order five or ten. In the latter case $G$ is the cyclic group $C_{10}$. Otherwise choose $a$ of order five. Its cyclic subgroup $H=⟨a⟩$ has index two and is therefore a normal subgroup.

In the noncyclic case an element $b$ outside $H$ has order two: its order divides ten, it cannot have order ten, and an element of order five has trivial image in the order-two quotient group $G/H$. Every element of $G$ is then $a^j$ or $a^{j}b$. Conjugation by $b$ restricts to an automorphism of $H$, say $ba{b}^{-1}=a^r$. Since $b^2=1$, $r^2\equiv 1(\mathrm{mod}5)$, so $r=1$ or $-1$ modulo five.

For $r=1$, the generators commute and $ab$ has order ten, yielding the cyclic case. For $r=-1$, these are the relations of the dihedral group of the pentagon. The classification of groups of order ten is therefore exactly $C_{10}$ and $D_{10}$, where the subscript on $D_{10}$ denotes its order. They are not isomorphic because one is Abelian and the other is not.

It suffices to centralize each minimal normal subgroup $M$. The subgroup $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first caseIn the second case $M$ is a nilpotent group. Its nontrivial group center is a characteristic subgroup of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are characteristic subgroups, and the subgroup killed by a prime show that $M$ is elementary abelian of exponent $p$. Thus $M\le O_p(G)$.

The finite p-group $O_p(G)$ acts by conjugation on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne 1$. This intersection is normal in $G$, and minimality makes it $M$. The other p-cores commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the group socle is generated by these minimal normal subgroups,

A soluble group is a group whose derived series $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$ terminates at $1$. Equivalently, it has a finite subnormal series with abelian factors. For the reverse implication, if $G/N$ has derived length $r$, then $G^{(r)}\le N$; if $N$ has derived length $s$, then $G^{(r+s)}=1$. Iterating this observation through the series proves the equivalence. For a finite group, refining the abelian factors to composition factors says that solubility is equivalent to all composition factors being cyclic of prime order.

Subgroups and quotient groups of soluble groups are soluble, since their derived series lie in, or are images of, the original series. The same calculation proves closure under group extensions. Finite p-groups are nilpotent groups, hence soluble: their nontrivial centers allow induction to construct a central series. Every finite nilpotent group is a direct product of groups of its Sylow subgroups. The converse fails: $S_3$ has the abelian-factor series $1<C_3<S_3$, but its order-two Sylow subgroups are not normal. The symmetric group $S_4$ is also soluble, with series $1<V_4<A_4<S_4$ and abelian factors. A finite soluble group has elementary abelian minimal normal subgroups, by the argument in 1(c). In contrast, a nonabelian simple group is a perfect group and is not soluble.

The Fitting subgroup collects the normal nilpotent structure. For a nontrivial finite soluble group it is nontrivial, since it contains an elementary abelian minimal normal subgroup. Part 1(c) proves $C_G(F)\le F$, so conjugation embeds $G/C_G(F)$ in $\mathrm{Aut}(F)$. This shows why controlling a soluble group through normal abelian or nilpotent layers is effective, even when the whole group is not a nilpotent group.

For a set of primes $\pi$, a Hall subgroup of type $\pi$ has order divisible only by primes in $\pi$ and index divisible only by primes outside $\pi$. The Hall theorem for soluble groups says that a finite soluble group has such subgroups, any two are conjugate, and every subgroup of $\pi$-order is contained in one. These assertions generalize the Sylow theorems, which treat a single prime. Here is a proof of all three assertions.

First we need coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary abelian p-group and $Q=E/V$ has order $h$ prime to $p$. A section $s:Q\to E$ defines an action on $V$ and an additive factor set $f(x,y)$ by $s(x)s(y)=f(x,y)s(xy)$. Associativity saysSumming over $z\in Q$ and dividing by $h$ in $V$ gives $f(x,y)=b(x)+xb(y)-b(xy)$, where $b(x)=h^{-1}{\sum }_{z}f(x,z)$. Replacing $s(x)$ by $-b(x)s(x)$ removes the factor set, giving a complement. Two complements differ by a map $d$ satisfying $d(xy)=d(x)+xd(y)$. Averaging over $y$ gives $d(x)=b-xb$, so the complements are conjugate by an element of $V$. This proves both splitting and conjugacy, without assuming that an arbitrary extension splits.

Induct on $|G|$, taking an elementary abelian minimal normal subgroup $V$ of exponent $p$. Choose a quotient Hall subgroup $\bar{H}\le G/V$ and let $K$ be its preimage. If $p\in \pi$, then $K$ itself is a Hall subgroup. If $p\notin \pi$, the preceding splitting argument gives a complement $H$ to $V$ in $K$, and that complement is a Hall subgroup of $G$.

For conjugacy, the images of two Hall subgroups are quotient Hall subgroups, so induction allows them to have the same image and lie in the same $K$. If $p\in \pi$, every Hall subgroup contains $V$: its product with $V$ is a $\pi$-subgroup and cannot be larger by the subgroup-order formula. Both are then $K$. If $p\notin \pi$, both are complements to $V$ in $K$, so the averaging argument conjugates them.

Finally let $U\le G$ be any $\pi$-subgroup. By induction its image lies in a quotient Hall subgroup, so $U\le K$. If $p\in \pi$, $K$ is already a Hall subgroup containing $U$. Otherwise write $K=V⋊H$. In $VU$, both $U$ and the inverse image in $H$ of the image of $U$ are complements to $V$. Complement conjugacy therefore places $U$ in a conjugate of $H$. This proves embedding and completes the induction.

Thus existence, conjugacy and containment all hold for every prime set in a finite soluble group. The solubility hypothesis matters: $A_5$ has no Hall subgroup for $\pi =\{2,5\}$. Such a subgroup would have index three; its coset action would give a nontrivial group homomorphism $A_5\to S_3$, impossible because $A_5$ is a simple group and has order $60$. This example also distinguishes a Hall subgroup from a merely arbitrary subgroup whose order uses the chosen primes.

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

The orbits of $N_{\alpha }⊴G_{\alpha }$ on $\Omega \setminus \{\alpha \}$ have a common size $r$, because $G_{\alpha }$ is transitive there. Suppose $N$ has a block system of block size $b$, $1<b<n$. Since $r\mid n-1$ and $b\mid n$, $\gcd (r,b)=1$. Let $B$ be the block of $\alpha$ and $C\ne B$ another block. The union $N_{\alpha }C$ is a union of blocks and of r-element point-orbits, so its size is divisible by $br$; it is at most $br$. Equality means each of the b point-orbits meets $C$ once. Thus $N_{\alpha \beta }$ fixes $C$ pointwise for $\beta \in C$. Interchanging $\alpha ,\beta$ also makes it fix $B$. All two-point stabilizers have order $|N_{\alpha }|/r$. Therefore $N_{\alpha \beta }=N_{\alpha \gamma }$ for every $\gamma \in B\setminus \{\alpha \}$. Varying $\beta$ outside $B$ shows this same subgroup fixes every point, so it is trivial. We have proved uniform subdegrees force an imprimitive action to be Frobenius: no nonidentity element of $N$ fixes two points.

Write $|N_{\alpha }|=h$. The distinct point stabilizers meet only at $1$, and $|N|=nh$. Thus the set $D$ of fixed-point-free elements of $N$ has sizeCount triples $(\alpha ,\beta ,x)$ with $x\in D$ and $x\alpha =\beta$: there are $n(n-1)$, exactly the number of ordered distinct pairs. Conjugation by the two-transitive group makes every pair occur, hence each pair has a unique such $x$. It follows that $G$ conjugates all elements of $D$ transitively.

Also $h\mid n-1$, since $N_{\alpha }$ acts freely on the remaining points. For each prime $p\mid n$, Cauchy's theorem for finite groups supplies an order-p element of $N$; it cannot fix a point, since $p\nmid h$. Conjugacy of $D$ forces all these primes to coincide, so $n=p^d$ and every element of $D$ has order $p$. A Sylow subgroup $P$ of $N$ has order $n$, because $p\nmid h$. Every nonidentity element of $P$ is fixed-point-free, so $P=\{1\}\cup D$. This uniquely described subgroup is normal in $G$. Minimal normality of $N$ gives $P=N$, and $|N|=n$. Therefore $N$ is regular. The proof establishes closure through a Sylow subgroup; it does not assume that a set of derangements is automatically a subgroup.

A finite minimal normal subgroup is a characteristically simple group, and has the direct-product structure of a finite minimal normal subgroupwith isomorphic simple groups $T_i$. For completeness, choose a minimal nontrivial normal subgroup $T$ of $N$. Its distinct $G$-conjugates are minimal normal subgroups of $N$, so they centralize one another. Their product is normal in $G$, hence equals $N$. If $T$ is abelian the product is abelian and minimal primary characteristic subgroups make $N$ elementary abelian. Otherwise each $T$ has trivial center: its center is a characteristic subgroup of $T$, hence normal in $N$, and minimality would make $T$ abelian. Each new conjugate then meets the product of previous conjugates trivially, because it centralizes that product and is centerless. The product is direct. Every normal subgroup of a direct factor is normal in $N$, so minimality of that factor makes it simple.

Assume now that $N$ acts primitively and nonregularly. Any nontrivial normal subgroup of this faithful primitive action is transitive. Thus each $T_i$ is transitive. If $k\ge 2$, the transitive commuting groups $T_1$ and $T_2$ are regular: an element of one fixing $\alpha$ commutes with the other and hence fixes its entire orbit. But the centralizer of a regular group has order $n$, as its elements are determined by the image of $\alpha$. Therefore all remaining factors lie in the order-n centralizer of $T_1$, and the transitive group $T_2$ already fills it. There can be only two factors, with $|N|=n^2$. Now $|N_{\alpha }|=n$, whereas its common nontrivial subdegree $r$ divides both $n$ and $n-1$. Hence $r=1$, making $N_{\alpha }$ trivial, a contradiction. Thus $k=1$, and $N$ is simple. An abelian transitive group is regular, so $N$ is nonabelian.

Finally $C_G(N)$ is normal in the primitive group $G$. If nontrivial it would be transitive, and commuting with transitive $N$ would force $N$ regular. Thus $C_G(N)=1$, and conjugation embeds $G$ in $\mathrm{Aut}(N)$. The centerless simple group $N$ is identified with its inner automorphism group. Therefore

The orbits of a normal subgroup $N$ are blocks for the primitive group action. If every orbit is a singleton then $N\le G_{[\Omega ]}$, giving the second alternative. Otherwise $N$ is transitive, and $G=N{G}_{\alpha }$.

Work in $G/N$. Every $g\in G$ can be written $nh$ with $n\in N$, $h\in G_{\alpha }$. Conjugation by $n$ has no effect on the image of a subgroup in this quotient, while $h$ normalizes $A$. Thus every conjugate of $A$ has the same image $AN/N$. The generation hypothesis says this image generates all of $G/N$. But it is already an abelian subgroup, so $G/N$ is abelian and $G^{\prime }\le N$. This proves the Iwasawa simplicity lemma in its possibly nonfaithful form:

Let $S=S{L}_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has kernel exactly $Z$: a linear map fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent vectors and extended to a basis. A linear map between two such bases can be adjusted to determinant one by scaling one target basis vector; this preserves both target lines. Therefore $S/Z$ has a faithful primitive group action.

Fix $L=Fv$. The mapsform an abelian subgroup: all products of the rank-one terms vanish. It is normal in the line stabilizer, since conjugation replaces $v$ by a scalar multiple and transports $f$. Its conjugates contain every transvection, and hence generate $S$ by 4(b). The same statements hold for its image in $S/Z$.

It remains to prove perfectness rather than assume it. Use the group commutator convention $[x,y]=xy{x}^{-1}{y}^{-1}$. If $n\ge 3$, distinct $i,j,k$ giveEvery elementary generator is consequently in $S^{\prime }$. If $n=2$ and $|F|>3$, choose $a\in F^{\ast }$ with $a^2\ne 1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\mathrm{diag}(a,a^{-1})$,Varying $b$ gives every upper transvection; conjugating by $w(1)$ gives every lower one. Again $S^{\prime }=S$, and its projective quotient is a perfect group.

Apply 4(a) to this faithful primitive projective action. Every nontrivial normal subgroup contains the commutator subgroup, which is the whole quotient. ThusThe restriction is real: $PS{L}_2(2)\cong S_3$ and $PS{L}_2(3)\cong A_4$ are not simple.

Work with finite groups, as required for these Sylow subgroup counts. Let $P$ act by conjugation on the set of Sylow $p$-subgroups. The stabilizer of $Q$ is $P\cap N_G(Q)$. Since $Q$ is the normal Sylow subgroup of its own normalizer, every $p$-subgroup of that normalizer lies in $Q$. Consequently the stabilizer is exactly $P\cap Q$, and the orbit-stabilizer theorem gives orbit size $[P:P\cap Q]$. The orbit of $P$ has size one. Every other orbit size is a power of $p$ at least $p^a$, hence is divisible by $p^a$. Adding orbit sizes proves the strengthened Sylow congruence from intersections:Now suppose $G$ is simple of order $2^e\cdot 15$. If $e=0$, its Sylow $5$-subgroup is normal because its number divides three and is one modulo five. If $e=1$, an involution in the regular permutation action of $G$ swaps fifteen pairs, so its sign is negative. The sign of a permutation would give a nontrivial group homomorphism $G\to C_2$ with a proper nontrivial normal kernel. Both cases contradict simplicity. Thus $e\ge 2$.

The Sylow theorems make $n_2$ a divisor of fifteen. It is not one, since a nontrivial Sylow subgroup would then be normal. It is not three, since the conjugation action on three Sylow subgroups would give a faithful group homomorphism $G\hookrightarrow S_3$, impossible by order. If $n_2=5$, take a Sylow subgroup $P$ itself: its normalizer has subgroup index five. If $n_2=15$, not all distinct Sylow subgroups can intersect $P$ in subgroup index at least four, since that would give $15\equiv 1(\mathrm{mod}4)$. Choose $Q\ne P$ with $[P:P\cap Q]=2$. Put $R=P\cap Q$, of order $2^{e-1}$. It has subgroup index two in both $P,Q$, so is normal in each. Hence $N_G(R)$ contains both distinct Sylow subgroups and has order greater than $2^e$. It is proper: otherwise $R$ would be a nontrivial proper normal subgroup of $G$. Its order is therefore $2^e\cdot 3$ or $2^e\cdot 5$. The latter would give an index-three subgroup and again an impossible faithful action on three cosets. Thus $[G:N_G(R)]=5$.

In either case we have the requested subgroup of order $2^e$ or $2^{e-1}$ with index-five normalizer. The action on the five cosets of that normalizer is nontrivial and faithful by simplicity. Its image lies in $A_5$, since composing with sign cannot give a nontrivial map from this simple group of order greater than two to $C_2$. Thus $2^e\cdot 15$ divides $60$, forcing $e=2$ and equality of orders. This proves the simple groups with order a power of two times fifteen conclusion:For $S{L}_2(5)$, choose the first column: there are $5^2-1=24$ nonzero choices. For each, determinant one is a nonzero linear equation in the second column and has five solutions. Therefore $|S{L}_2(5)|=120$. A central matrix commuting with $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ has zero lower-left entry and equal diagonal entries; commuting also with $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ makes the upper-right entry zero. It is scalar. Determinant one then givesTo establish the projective special linear group over the field with five elements isomorphism without assuming its simplicity, compute its classes. For determinant-one matrices, the characteristic polynomial is $X^2-tX+1$ with trace $t$. The possibilities over ${\mathbb{F}}_5$ exhaust the group:For trace zero the eigenvalues are $2,3$, so the determinant-one centralizer is the split diagonal torus of order four. For traces $\pm 1$, the discriminant is the nonsquare $2$: the centralizer is the norm-one subgroup of ${\mathbb{F}}_{25}^{\times }$, of order six. In both semisimple cases the determinant map on the full linear centralizer is surjective onto ${\mathbb{F}}_5^{\times }$; consequently each trace type is a single special-linear class. For the noncentral traces $\pm 2$, a Jordan block has full linear centralizer $\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right):a\ne 0\right\}$ with determinant $a^2$. The determinant image consists of the two squares, so each general-linear class splits into two special-linear classes; the determinant-one centralizer has order $2\cdot 5=10$. These sizes sum to $120$.

Quotienting by $Z$ identifies a matrix with its negative. The trace-zero class gives one projective class of size fifteen; the two traces $\pm 1$ give one class of size twenty; the four noncentral unipotent classes give two classes of size twelve. Their respective projective orders are $2,3,5,5$, by the characteristic polynomials. A normal subgroup is a union of classes containing the identity. The nontrivial proper possible union sizes are $13,16,21,25,28,33,36,40,45,48$, none of which divides $60$. Thus the quotient is simple. Applying the order-$60$ result above givesFinally, if $A^2=I$ over an odd-characteristic field, its minimal polynomial divides the square-free polynomial $(X-1)(X+1)$, so it is diagonalizable with eigenvalues $\pm 1$. In dimension two, determinant one requires both eigenvalues equal. Hence $A=I$ or $A=-I$. This proves the unique involution in SL2 over an odd field assertion: the only element of order two in $S{L}_2(5)$ is $-I$. If an index-two subgroup $H$ existed, it would be normal and have order sixty. By Cauchy theorem it would contain an involution, hence contain $Z$. Then $H/Z$ would be a normal subgroup of order thirty in the simple quotient $PS{L}_2(5)$, a contradiction. Thus $S{L}_2(5)$ has no index-two subgroup.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.