= Solution
A <soluble group> is a <group> whose <derived series> $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$ terminates at $1$. Equivalently, it has a finite <subnormal series> with abelian factors. For the reverse implication, if $G/N$ has derived length $r$, then $G^{(r)}\le N$; if $N$ has derived length $s$, then $G^{(r+s)}=1$. Iterating this observation through the series proves the equivalence. For a finite <group>, refining the abelian factors to <composition factors> says that solubility is equivalent to all <composition factors> being cyclic of prime order.
<Subgroups> and <quotient groups> of <soluble groups> are soluble, since their <derived series> lie in, or are images of, the original series. The same calculation proves closure under <group extensions>. <Finite p-groups> are <nilpotent groups>, hence soluble: their nontrivial centers allow induction to construct a <central series>. Every finite <nilpotent group> is a <direct product of groups> of its <Sylow subgroups>. The converse fails: $S_3$ has the abelian-factor series $1<C_3<S_3$, but its order-two <Sylow subgroups> are not normal. The <symmetric group> $S_4$ is also soluble, with series $1<V_4<A_4<S_4$ and abelian factors. A finite <soluble group> has elementary abelian <minimal normal subgroups>, by the argument in 1(c). In contrast, a nonabelian <simple group> is a <perfect group> and is not soluble.
The <Fitting subgroup> collects the normal nilpotent structure. For a nontrivial finite <soluble group> it is nontrivial, since it contains an elementary abelian <minimal normal subgroup>. Part 1(c) proves $C_G(F)\le F$, so <conjugation> embeds $G/C_G(F)$ in $\operatorname{Aut}(F)$. This shows why controlling a <soluble group> through normal abelian or nilpotent layers is effective, even when the whole <group> is not a <nilpotent group>.
For a set of primes $\pi$, a <Hall subgroup> of type $\pi$ has order divisible only by primes in $\pi$ and index divisible only by primes outside $\pi$. The <Hall theorem for soluble groups> says that a finite <soluble group> has such <subgroups>, any two are conjugate, and every <subgroup> of $\pi$-order is contained in one. These assertions generalize the <Sylow theorems>, which treat a single prime. Here is a proof of all three assertions.
First we need <coprime splitting over an elementary abelian normal subgroup>. Suppose $V\trianglelefteq E$ is an elementary abelian p-group and $Q=E/V$ has order $h$ prime to $p$. A section $s:Q\to E$ defines an action on $V$ and an additive factor set $f(x,y)$ by $s(x)s(y)=f(x,y)s(xy)$. Associativity says
$$
f(x,y)+f(xy,z)=x f(y,z)+f(x,yz).
$$
Summing over $z\in Q$ and dividing by $h$ in $V$ gives $f(x,y)=b(x)+xb(y)-b(xy)$, where $b(x)=h^{-1}\sum_zf(x,z)$. Replacing $s(x)$ by $-b(x)s(x)$ removes the factor set, giving a complement. Two complements differ by a map $d$ satisfying $d(xy)=d(x)+xd(y)$. Averaging over $y$ gives $d(x)=b-xb$, so the complements are conjugate by an element of $V$. This proves both splitting and conjugacy, without assuming that an arbitrary extension splits.
Induct on $|G|$, taking an elementary abelian <minimal normal subgroup> $V$ of exponent $p$. Choose a quotient <Hall subgroup> $\bar H\le G/V$ and let $K$ be its preimage. If $p\in\pi$, then $K$ itself is a <Hall subgroup>. If $p\notin\pi$, the preceding splitting argument gives a complement $H$ to $V$ in $K$, and that complement is a <Hall subgroup> of $G$.
For conjugacy, the images of two <Hall subgroups> are quotient <Hall subgroups>, so induction allows them to have the same <image> and lie in the same $K$. If $p\in\pi$, every <Hall subgroup> contains $V$: its product with $V$ is a $\pi$-subgroup and cannot be larger by the subgroup-order formula. Both are then $K$. If $p\notin\pi$, both are complements to $V$ in $K$, so the averaging argument conjugates them.
Finally let $U\le G$ be any $\pi$-subgroup. By induction its <image> lies in a quotient <Hall subgroup>, so $U\le K$. If $p\in\pi$, $K$ is already a <Hall subgroup> containing $U$. Otherwise write $K=V\rtimes H$. In $VU$, both $U$ and the inverse <image> in $H$ of the <image> of $U$ are complements to $V$. Complement conjugacy therefore places $U$ in a conjugate of $H$. This proves embedding and completes the induction.
\b[Thus existence, conjugacy and containment all hold for every prime set in a finite <soluble group>.] The solubility hypothesis matters: $A_5$ has no <Hall subgroup> for $\pi=\{2,5\}$. Such a <subgroup> would have index three; its coset action would give a nontrivial <group homomorphism> $A_5\to S_3$, impossible because $A_5$ is a <simple group> and has order $60$. This example also distinguishes a <Hall subgroup> from a merely arbitrary <subgroup> whose order uses the chosen primes.
Back to article page
