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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.

For subgroups $A,B$ write $[A,B]$ for the subgroup generated by their group commutators. A nilpotent group has a central series: a finite chain $1=G_0\le G_1\le \cdots \le G_c=G$ with $[G_i,G]\le G_{i-1}$. Equivalently its upper central series, defined by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, reaches $G$. Indeed every central series term $G_i$ lies in $Z_i$ by induction, and the upper central series itself is central. A third equivalent definition is that the lower central series ${\gamma }_1=G$, ${\gamma }_{i+1}=[{\gamma }_i,G]$ terminates at ${\gamma }_{c+1}=1$: reversing this series gives a central series, and induction along any central series gives ${\gamma }_{c+1}=1$.

The least such $c$ is the nilpotency class. Nontrivial abelian groups have class one. Upper-unitriangular $3\times 3$ matrices have class two: their group commutators lie in the central top-right-entry subgroup, and further group commutators vanish. The same example works over a finite field or over the integers, so nilpotence is not a finiteness property. Subgroups and quotients inherit nilpotence by intersecting or mapping a central series; finite direct products of groups have class at most the largest factor class. Every nilpotent group is soluble, since its derived series is contained successively in the lower central series. The converse fails: $S_3$ is soluble but its center is trivial, so it cannot be a nontrivial nilpotent group. Extensions of nilpotent groups need not be nilpotent, as the extension $C_3◃S_3$ with quotient $C_2$ shows.

To prove the finite classification, first note that every nontrivial finite p-group has nontrivial center. In its class equation, every noncentral conjugacy class has size divisible by $p$, so the center's order is divisible by $p$. Induct on the group order: the smaller p-group $P/Z(P)$ is nilpotent, and lifting its central series and adjoining $Z(P)$ proves $P$ nilpotent. A finite direct product of groups of prime-power-order groups is consequently nilpotent.

Conversely suppose $G$ is finite and nilpotent. We need the normalizer condition for nilpotent groups. For any proper subgroup $H<G$, choose the first upper-central term $Z_i$ not contained in $H$. Then $Z_{i-1}\le H$. Every $z\in Z_i$ commutes with $H$ modulo $Z_{i-1}$, so it normalizes $H$. Choosing $z\notin H$ proves $H<N_G(H)$.

Let $P$ be a Sylow subgroup and put $M=N_G(P)$. Since $P$ is normal in $M$, it is the unique Sylow p-subgroup of $M$. Any element normalizing $M$ must therefore normalize $P$, so $N_G(M)=M$. If $M$ were proper, the normalizer condition would contradict this equality. Hence $N_G(P)=G$: every Sylow subgroup is normal. For Sylow subgroups $P,Q$ belonging to different primes, $[P,Q]\le P\cap Q=1$, so they commute. Their product has order the product of their orders, namely $|G|$, and has trivial intersections between each factor and the product of the others. Multiplication therefore givesThis proves finite nilpotent group decomposition in both directions without assuming the classification in the normalizer argument. It also shows that a finite group is nilpotent exactly when all its Sylow subgroups are normal. The finite hypothesis is essential to the prime-power direct-product statement; an infinite cyclic group is nilpotent but is not such a finite product.