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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.