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A Hall subgroup has order coprime to its index. For a set of primes $\pi$, a Hall $\pi$-subgroup has order using only primes of $\pi$ and index using only primes outside $\pi$. Thus its order contains the full prime-power contributions belonging to $\pi$. A Sylow subgroup is the single-prime instance. Hall subgroup existence in soluble groups generalizes Sylow existence to every prime set.

A soluble group is one whose derived series terminates: $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, and $G^{(r)}=1$ for some finite $r$. Each factor of this series is abelian. Conversely, if a subnormal series has abelian factors, repeated commutation moves down the series, so the derived series terminates. This explains the usual alternative definition in terms of abelian factors. The least $r$ is the derived length.

Subgroups are soluble because $H^{(i)}\le G^{(i)}$; quotients are soluble because a group homomorphism maps a derived subgroup onto the derived subgroup of its image. Solubility is also closed under extensions: if $N◃G$ has derived length at most $a$ and $G/N$ has length at most $b$, then $G^{(b)}\le N$ and $G^{(a+b)}=1$. Abelian and nilpotent groups are soluble; $S_3$ has derived subgroup $A_3$ and derived length two. In a finite group, solubility is equivalent to every composition factor being cyclic of prime order: simple soluble groups are abelian and therefore cyclic of prime order, while a composition series with these factors builds a soluble group by extensions. This need not give an ambient-normal series with cyclic prime-order factors, which is the stronger property of supersolubility.

A Hall subgroup for a prime set $\pi$ is a subgroup $H$ whose order uses only primes in $\pi$ and whose index uses only primes outside $\pi$. Its order is the full $\pi$-part of $|G|$. Sylow's theorem is the singleton-prime case; solubility allows all prime sets, although Hall subgroups need not be normal, as the order-two subgroups of $S_3$ show.

Prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$. It is elementary abelian of some prime characteristic $p$: $N^{\prime }$ is characteristic in $N$ and normal in $G$, so minimality and solubility force $N^{\prime }=1$. A nontrivial Sylow subgroup of this finite abelian group is characteristic, making $N$ a p-group. The subgroup of pth powers is characteristic and proper, so minimality makes it trivial. This proves minimal normal subgroups of finite solvable groups are elementary abelian.

By induction $G/N$ has a Hall $\pi$-subgroup $\overline{H}$. Let $K$ be its inverse image. Then $[G:K]$ is a ${\pi }^{\prime }$-number. If $p\in \pi$, the whole $K$ is a $\pi$-group and is the desired Hall subgroup. If $p\notin \pi$, $K/N=Q=\overline{H}$ has order $s$ prime to $p$. We now prove the required complement existence explicitly, rather than invoking a splitting theorem as a substitute.

Write $N$ additively as an ${\mathbb{F}}_p$-vector space. Choose representatives $u(x)$ for $x\in Q$ with $u(1)=1$. Conjugation gives a well-defined action of $Q$ on $N$, because changing a representative by an element of the abelian group $N$ does not change that action. Define $f(x,y)\in N$ by $u(x)u(y)=f(x,y)u(xy)$. Associativity yieldsSince $s$ is invertible in ${\mathbb{F}}_p$, set $c(x)=s^{-1}{\sum }_{z\in Q}f(x,z)$. Sum the identity over $z$ and use the bijection $z\mapsto yz$ to obtainChanging representatives to $u^{\prime }(x)=(-c(x))u(x)$ removes the multiplication error:Thus $x\mapsto u^{\prime }(x)$ is a homomorphic section. Its image $H$ has order $|Q|$, intersects $N$ trivially, and complements $N$ in $K$. This is coprime splitting over an elementary abelian normal subgroup. Finally $|H|$ is a $\pi$-number and $[G:H]=[G:K]|N|$ is a ${\pi }^{\prime }$-number, soOnly existence was needed here; the fuller Hall theorem also gives conjugacy and containment of arbitrary $\pi$-subgroups. Solubility makes the elementary-abelian minimal-normal induction possible and cannot simply be omitted.

For a set of primes $\pi$, a Hall $\pi$-subgroup $H$ of a finite group is a subgroup whose order uses only primes in $\pi$ and whose subgroup index uses only primes outside $\pi$. Equivalently, $|H|$ is the full $\pi$-part of $|G|$. A Hall subgroup is a Hall $\pi$-subgroup for some $\pi$. The theorem of P. Hall for finite soluble groups states: Hall $\pi$-subgroups exist for every $\pi$, any two are conjugate, and every $\pi$-subgroup is contained in a Hall $\pi$-subgroup.

For the existence proof, first give the explicit coprime splitting calculation needed in the induction. Let $N$ be elementary abelian of field characteristic $r$, normal in $E$, with $r\nmid |E/N|$. Write $Q=E/N$ and use additive notation on $N$. A normalized section $s:Q\to E$ gives an action $x\cdot n=s(x)ns(x{)}^{-1}$ independent of its choice, since $N$ is abelian. Write $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in Q$, put $S(x)={\sum }_{z}f(x,z)$, and divide by $|Q|$ in the vector space $N$. This is legitimate because $r\nmid |Q|$. With $b(x)=|Q{|}^{-1}S(x)$ one obtainsThe new section $s^{\prime }(x)=(-b(x))s(x)$ therefore has zero multiplication defect and is a group homomorphism. Its image is a complement to $N$.

We will also need a single conjugation carrying one complement to the other. If two homomorphic sections differ by $d(x)$, their difference satisfies $d(xy)=d(x)+x\cdot d(y)$. Summing over $y$ gives $d(x)=c-x\cdot c$, where $c=|Q{|}^{-1}{\sum }_{y}d(y)$. Thus the second section is $cs(x)c^{-1}$. This proves, by explicit averaging, both coprime splitting over an elementary abelian normal subgroup and the existence of such a conjugation by an element of $N$.

Now induct on $|G|$ for Hall subgroup existence in soluble groups. A minimal normal subgroup $N$ of a finite soluble group is elementary abelian of some field characteristic $r$. Indeed, $N^{\prime }$ is a characteristic subgroup of $N$ and normal in $G$; it cannot equal $N$ because $N$ is soluble, so it is trivial. In this abelian group, a nontrivial primary component is a characteristic subgroup, forcing a single prime, and the characteristic subgroup annihilated by that prime forces exponent $r$.

By induction $G/N$ has a Hall $\pi$-subgroup $\overline{H}$, with preimage $M$. If $M<G$, induction in $M$ supplies a Hall $\pi$-subgroup of $M$, which is also Hall in $G$ because $[G:M]$ is prime to all primes in $\pi$. If $M=G$, the quotient is a $\pi$-group. When $r\in \pi$, $G$ itself is a $\pi$-group. When $r\notin \pi$, the averaging construction splits off $N$, and its complement has order $|G/N|$ and is the required Hall $\pi$-subgroup. This completes the existence proof, including the identity group and empty prime set.

A Sylow basis is a family $\{P_p:p\mid |G|\}$ containing one Sylow subgroup for each prime, with $P_p{P}_q=P_q{P}_p$ for each pair. This is permutability of subgroups, not commutation of all their elements. The product of any subfamily is a subgroup: commuting the factors as sets proves closure and closure under inversion. Inductively its order is the product of the prime-power orders, since the newly adjoined factor has trivial intersection with the earlier product. It is thus the corresponding Hall subgroup.

Prove existence of a Sylow basis by induction on $|G|$, again using elementary abelian $N$ of field characteristic $r$. Choose a basis $\{{\overline{P}}_p\}$ of $G/N$, taking ${\overline{P}}_r=1$ if $r$ does not divide that quotient's order. The product $\overline{H}={\prod }_{p\ne r}{\overline{P}}_p$ is its Hall $r^{\prime }$-subgroup. In its full preimage $T$, the averaging lemma supplies a complement $H$ to $N$, mapping isomorphically to $\overline{H}$. For $p\ne r$, let $P_p$ be the inverse image of ${\overline{P}}_p$ under this isomorphism within $H$. Let $P_r$ be the full preimage of ${\overline{P}}_r$ in $G$. These are Sylow subgroups of $G$. The other-prime pairs permute because they do so in $\overline{H}$. For $P_r$ and $P_p$, their product is contained in the full preimage of the subgroup ${\overline{P}}_r{\overline{P}}_p$ and has exactly that preimage's order $|N||{\overline{P}}_r||{\overline{P}}_p|$. Equality follows, so the product is a subgroup and these two factors permute as well. This constructs a basis.

To obtain a simultaneous conjugation of two bases, take two bases. Their images in $G/N$ are bases: $N$ lies in every Sylow $r$-subgroup, and the other-prime Sylow subgroups project injectively. By induction, conjugate one basis so that all corresponding images agree. The products of their non-$r$ factors are then two complements to $N$ in the same preimage $T$ of their common Hall $r^{\prime }$-subgroup. The averaging lemma for complements supplies a single element of $N$ conjugating these complements. In the resulting common complement each corresponding Sylow subgroup is the unique lift of its agreed quotient subgroup, so all non-$r$ factors agree. The $r$-factors already equal the full preimage of their common quotient factor and are preserved by conjugation from $N$. Thus all Sylow bases of a finite soluble group are conjugate.

In $S_4$ there are three Sylow $2$-subgroups of order eight: the action on the three pair-partitions of four points has kernel $V_4$, and these subgroups are the preimages of the three order-two subgroups of $S_3$. There are four Sylow $3$-subgroups, one for each three-point support. Every choice $P_2,P_3$ has trivial intersection and product of cardinality $8\cdot 3=24$, so $P_2{P}_3=S_4=P_3{P}_2$. ThereforeFor $E=V⋊S_4$, $V={\mathbb{F}}_5^4$, the Sylow $5$-subgroup is uniquely $V$. In a Sylow basis the product $P_2{P}_3$ is a Hall $5^{\prime }$-subgroup, hence a complement to $V$. Conversely, a basis of any complement together with $V$ is a basis of $E$, because $V$ is normal. The averaging lemma makes all complements conjugate by $V$. The stabilizer of the standard complement $S_4$ under this conjugation is $C_V(S_4)$: if a vector normalizes it, its group commutator with each complement element lies in both $V$ and that complement, hence is trivial. The fixed vectors under the natural coordinate permutation action are exactly $(a,a,a,a)$, so $|C_V(S_4)|=5$. There are consequently $5^4/5=125$ complements, and each has twelve bases. The complement is uniquely recovered from its basis as $P_2{P}_3$, so no basis is counted twice. The counting Sylow bases in a coprime elementary abelian extension formula yields

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.