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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