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The Hall theorem for soluble groups says that for every prime set $\pi$, a finite soluble group has Hall $\pi$-subgroups; they are all conjugate, and every $\pi$-subgroup is contained in one. A Hall $\pi$-subgroup has only primes in $\pi$ in its order and no primes in $\pi$ in its index.

For a proof outline, induct on the group order and choose a minimal normal subgroup $N$. In a finite soluble group this is elementary abelian of some characteristic $p$: its derived subgroup is characteristic and solvability forces it to be trivial; primary components and the subgroup of elements of order dividing $p$ are then characteristic, so minimality gives an elementary abelian p-group. Choose a Hall $\pi$-subgroup of $G/N$ and let $U$ be its preimage. If $p\in \pi$, $U$ is itself a Hall $\pi$-subgroup. If $p\notin \pi$, $U/N$ has order prime to $p$ and coprime splitting over an elementary abelian normal subgroup supplies a complement to $N$ in $U$, of the required Hall order.

Here the complement fact has a short averaging proof. Choose a section of $U/N=H$ and write its multiplication defect as $f(h,k)\in N$, in additive notation. Associativity givesSince $|H|$ is invertible on $N$, averaging over $t$ expresses the defect as $f(h,k)=b(h)+hb(k)-b(hk)$; adjusting the section by $-b$ gives a homomorphic section. Two such complements differ by $d(hk)=d(h)+hd(k)$. Averaging this identity gives $d(h)=b-hb$, so they are conjugate by $b\in N$. Inductive conjugacy in the quotient followed by this complement conjugacy proves Hall conjugacy. To embed an arbitrary $\pi$-subgroup $Q$, first put its quotient inside the chosen quotient Hall subgroup. In the coprime case $Q$ and the corresponding subgroup of a complement are both complements to $N$ inside $NQ$, hence are $N$-conjugate. In the other case $Q\le U$ directly. This proves Hall conjugacy and embedding in finite soluble groups as well as existence.

Now prove the assertion about maximal subgroups by induction on $|G|$. Suppose a nontrivial normal subgroup $K$ is contained in $A$. If $K\nleqq B$, maximality gives $KB=G$, and since $K\le A$ this gives $AB=G$. If $K\le B$, both $A/K$ and $B/K$ are maximal in $G/K$; induction yields either their product equal to $G/K$ or their conjugacy, lifting respectively to $AB=G$ or to conjugacy of $A,B$. The same applies with their roles reversed.

It remains to suppose that no nontrivial normal subgroup of $G$ lies in either maximum. Let $K$ be minimal normal, an elementary abelian p-group. Since $K$ lies in neither maximum, $KA=KB=G$. The intersection $K\cap A$ is normalized by $A$ and by the abelian group $K$, hence by $G$; it is therefore trivial. Likewise $K\cap B=1$. Thus both $A$ and $B$ complement $K$. Their action on $K$ is irreducible, since an invariant subspace would be a proper nontrivial normal subgroup of $G$. It is also faithful: $C_A(K)$ is normalized by $A$ and centralized by $K$, so is a normal subgroup of $G$ contained in $A$ and must be trivial.

If $G/K=1$, these assumptions force $A=B=1$ and $G$ to have prime order. Otherwise choose $L>K$ normal in $G$ with $L/K$ minimal normal in $G/K$. It is an elementary abelian r-group. Under $A\cong G/K$, its copy $R=A\cap L$ is a nontrivial normal subgroup of $A$. The prime $r$ differs from $p$: a p-group acting on the characteristic-p vector space $K$ has nonzero fixed points, by counting its orbits on $K$ modulo $p$. Its fixed subspace would be $A$-invariant, hence all of $K$ by irreducibility, contrary to faithfulness of $R$.

Therefore $A\cap L$ and $B\cap L$ are Sylow r-subgroups of $L=K(A\cap L)=K(B\cap L)$. They are conjugate in $L$. Conjugate $B$ so that these intersections coincide as $R$. Since $R⊲A$, $N_G(R)\ge A$. It cannot equal $G$, since $R$ would then be a nontrivial normal subgroup contained in $A$. Maximality gives $N_G(R)=A$, and the same argument gives $N_G(R)=B$. The conjugated maxima are equal. HenceThis is the factorization-or-conjugacy property of maximal subgroups of a finite soluble group.