= Solution
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 $H^{(i)}\leq G^{(i)}$; 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 $N\triangleleft G$ has derived length at most $a$ and $G/N$ has length at most $b$, then $G^{(b)}\leq N$ 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 $p$: $N'$ is characteristic in $N$ and normal in $G$, so minimality and solubility force $N'=1$. 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 $\pi'$-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 yields
$$
f(x,y)+f(xy,z)=x\cdot f(y,z)+f(x,yz).
$$
Since $s$ 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 obtain
$$
f(x,y)=c(x)+x\cdot c(y)-c(xy).
$$
Changing representatives to $u'(x)=(-c(x))u(x)$ removes the multiplication error:
$$
u'(x)u'(y)=u'(xy).
$$
Thus $x\mapsto u'(x)$ 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 $\pi'$-number, so
$$
\boxed{\text{every finite soluble }G\text{ has a Hall }\pi\text{-subgroup for every prime set }\pi.}
$$
Only 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.
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