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A Hall subgroup for a set of primes $\pi$ is a subgroup $H\le G$ such that every prime divisor of $|H|$ belongs to $\pi$, while no prime divisor of $[G:H]$ belongs to $\pi$. Equivalently $|H|=|G{|}_{\pi }$, the product of the full prime powers in $|G|$ belonging to $\pi$.

The Hall conjugacy and embedding in finite soluble groups theorem says that, for every $\pi$, a finite soluble group has a Hall $\pi$-subgroup, any two such subgroups are conjugate, and every $\pi$-subgroup is contained in one. We prove all three assertions together by induction, including the needed complement argument rather than assuming it.

First establish coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary Abelian $r$-group and $D=E/V$ has order $m$ prime to $r$. Write $V$ additively. Conjugation defines a well-defined action of $D$ on $V$ because $V$ is Abelian. Choose a section $s:D\to E$, with $s(1)=1$, and let its extension cocycle be defined by $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in D$. With $S(x)={\sum }_{z}f(x,z)$ this becomesMultiplication by $m$ is invertible on $V$, so setting $b(x)=m^{-1}S(x)$ gives $f(x,y)=b(x)+xb(y)-b(xy)$. The new section $s^{\prime }(x)=(-b(x))s(x)$ has zero cocycle and is a homomorphism. Its image is a complement $K$ to $V$, proving $E=V⋊K$.

Every other complement has the form $\{h(x)s^{\prime }(x):x\in D\}$, with $h(xy)=h(x)+xh(y)$. Summing this identity over $y$ and putting $T={\sum }_{y}h(y)$ gives $mh(x)=T-xT$. Hence $h(x)=v-xv$, where $v=m^{-1}T$, and the other complement is $vK{v}^{-1}$. The same argument applies on any subgroup $D_0\le D$: a subgroup intersecting $V$ trivially is a graph of such a one-cocycle over its image $D_0$, so conjugation by an element of $V$ carries it into $K$. This proves the complement-conjugacy and embedding assertions needed below.

Now let $G\ne 1$ be finite soluble, and choose a minimal normal subgroup $V$. It is elementary Abelian of some prime characteristic $r$, by the minimal-normal argument in the series discussion. Induction gives Hall subgroups and the embedding and conjugacy assertions in $G/V$. Let $\overline{H}$ be a Hall $\pi$-subgroup there and $M$ its full preimage.

If $r\in \pi$, then $M$ itself is a Hall $\pi$-subgroup of $G$. Every Hall $\pi$-subgroup $H$ contains $V$, since $HV$ is a $\pi$-subgroup and $H$ already has the largest possible $\pi$-order. Thus $H$ is the full preimage of its image in $G/V$, and quotient conjugacy lifts to conjugacy in $G$. Given any $\pi$-subgroup $U$, induction puts its quotient image into some $\overline{H}$, whose preimage $M$ contains $U$.

If $r\notin \pi$, then $M/V=\overline{H}$ has order prime to $r$. The complement construction produces $K\le M$ of order $|\overline{H}|=|G{|}_{\pi }$, hence a Hall subgroup of $G$. Any Hall subgroup has trivial intersection with $V$ and a Hall image in $G/V$. Conjugating two Hall subgroups first aligns these images, after which both are complements to $V$ in the same $M$; the complement-conjugacy argument then conjugates them by an element of $V$. For an arbitrary $\pi$-subgroup $U$, quotient induction puts its image into some $\overline{H}$. Then $U\subseteq M$ and $U\cap V=1$, so the embedding argument conjugates $U$ into the chosen complement $K$. Equivalently $U$ lies in a conjugate of $K$. This completes the induction and all three parts of the Hall theorem for soluble groups.

For the specific general linear group over a finite field, counting ordered bases givesThere are seven lines and seven planes in ${\mathbb{F}}_2^3$. The group is transitive on each family, so line and plane stabilizers both have order $24$ and index seven. To show that these exhaust the index-seven subgroups, let $H$ have order $24$. It contains a Sylow $2$-subgroup of order eight, which can be conjugated toThe group orbits of $U$ on the seven nonzero vectors areof sizes $1,2,4$. An $H$-orbit is a union of these. The group $H$ cannot be transitive on seven points, since seven does not divide $24$. Therefore it has an orbit of size at most three. A size-one orbit fixes $e_1$. A size-two orbit must be $\{e_2,e_1+e_2\}$, whose vector sum $e_1$ is consequently fixed. A size-three orbit must be the union of the first two $U$-orbits, the nonzero part of the plane $⟨e_1,e_2⟩$, so that plane is invariant. In every case $H$ lies in a line or plane stabilizer, and equality follows from their equal orders.

The two stabilizer families are not conjugate. A line stabilizer fixes its nonzero vector and is transitive on the other six nonzero vectors: extend that fixed vector and either chosen other vector to a basis, and map one such basis to the other. Thus it has no invariant three-element subset of nonzero vectors and fixes no plane. In contrast a plane stabilizer fixes its defining plane. Conjugation sends line stabilizers to line stabilizers, so cannot send one to a plane stabilizer. We obtainThey are Hall subgroups for $\pi =\{2,3\}$ but are not conjugate. This explicitly shows why the solubility hypothesis of the Hall theorem for soluble groups matters.

Finally, an index-three subgroup would give a nontrivial transitive coset homomorphism $G{L}_3(2)\to S_3$. Its kernel is normal; the permitted simplicity of $G{L}_3(2)$ forces that kernel to be trivial, because the action is nontrivial. But an injective map from a group of order $168$ into one of order six is impossible. Hence $G{L}_3(2)$ has no subgroup of index three.

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.