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The Frattini subgroup is the intersection of the maximal subgroups of $G$, with $\Phi (1)=1$. A useful equivalent property isIndeed, a proper subgroup $H$ lies in a maximal subgroup, which also contains $\Phi (G)$. The Frattini subgroup is a characteristic subgroup, since automorphisms permute the maximal subgroups.

The Fitting subgroup is the largest nilpotent normal subgroup. To prove its existence, first form the p-core $O_p(G)$: the product of all normal p-subgroups. For two such subgroups, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many subgroups, so $O_p(G)$ exists. For distinct primes $p,r$,Thus the product of the p-cores is a nilpotent normal subgroup, a direct product of groups of its Sylow subgroups. Conversely, a finite nilpotent normal subgroup $K$ has unique Sylow subgroups, each a characteristic subgroup of $K$. Each is normal in $G$ and lies in the corresponding p-core. Consequently the required largest subgroup exists, and

Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a minimal normal subgroup $K/D$ of $G/D$ contained in $C/D$. In a finite soluble group, such a minimal normal subgroup is elementary abelian: the last nontrivial term of its derived series is an abelian characteristic subgroup, so minimality makes the whole subgroup abelian; its primary components and subgroup of elements killed by $p$ then force one prime and exponent $p$.

It follows that $K^{\prime }\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K^{\prime }\le Z(K)$. Therefore $K$ is a nilpotent group of class at most two. Since $K$ is normal in $G$, the definition of the Fitting subgroup gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the Fitting subgroup is self-centralizing in soluble groups:

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 $\mathrm{Aut}(F)$. 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 $V⊴E$ 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 saysSumming over $z\in Q$ and dividing by $h$ in $V$ gives $f(x,y)=b(x)+xb(y)-b(xy)$, where $b(x)=h^{-1}{\sum }_{z}f(x,z)$. 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 $K=V⋊H$. 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.

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.