Hall subgroups exist in finite solvable

Statement
Suppose $$G$$ is a fact about::finite solvable group, and $$\pi$$ is a prime set. Then, there exists a $$\pi$$-fact about::Hall subgroup of $$G$$: a subgroup whose order and index are relatively prime, and with the property that the set of prime divisors of its order is within $$\pi$$.

Related facts

 * Hall's theorem is a converse of sorts.
 * Sylow's theorem: Sylow subgroups exist, Sylow implies order-conjugate, Sylow implies order-dominating, congruence condition on Sylow numbers
 * ECD condition for pi-subgroups in solvable groups: Hall subgroups exist in finite solvable, Hall implies order-conjugate in finite solvable, Hall implies order-dominating in finite solvable, congruence condition on factorization of Hall numbers
 * Pi-Hall subgroups exist in pi-separable

Facts used

 * 1) uses::Solvability is subgroup-closed
 * 2) uses::Solvability is quotient-closed
 * 3) uses::Minimal normal implies elementary Abelian in finite solvable
 * 4) uses::Sylow subgroups exist
 * 5) uses::Normality is quotient-transitive
 * 6) uses::Frattini's argument
 * 7) uses::Product formula

Proof
Given: A finite solvable group $$G$$, a prime set $$\pi$$.

To prove: $$G$$ has a $$\pi$$-Hall subgroup.

Proof: We prove the claim by induction on the order of $$G$$. The base case of the trivial group is clear.

Suppose the result is true for all finite solvable groups of order strictly less than the order of $$G$$. Then, by facts (1) and (2), the result is true for all proper subgroups and all proper quotients of $$G$$.

Let $$N$$ be a minimal normal subgroup of $$G$$. By fact (3), $$N$$ is elementary Abelian, and in particular, is a $$p$$-group for some prime divisor of $$G$$. Observe that:


 * 1) Case $$p \in \pi$$: By applying induction to $$G/N$$, we conclude that $$G/N$$ has a $$\pi$$-Hall subgroup. Taking the inverse image of this in $$G$$, we get a $$\pi$$-Hall subgroup of $$G$$.
 * 2) Case $$p \notin \pi$$: By applying induction to $$G/N$$, we conclude that $$G/N$$ has a $$\pi$$-Hall subgroup. The inverse image of this gives a subgroup, say $$K$$, of $$G$$. If $$G/N$$ is not itself a $$\pi$$-group, then $$K/N$$ is proper in $$G/N$$, and $$K$$ is a proper subgroup of $$G$$, then $$K$$ has a $$\pi$$-Hall subgroup $$H$$. Note that $$[G:K] = [G/N:K/N]$$ is relatively prime to $$\pi$$, and $$[K:H]$$ is relatively prime to $$\pi$$, so $$[G:H]$$ is relatively prime to $$\pi$$. Thus, $$H$$ is a Hall $$\pi$$-subgroup of $$G$$.

Thus, the only case of concern is where $$p \notin \pi$$ but $$G/N$$ is a $$\pi$$-group. Note that in this case, $$N$$ must be a $$p$$-Sylow subgroup (since $$G/N$$ has order relatively prime to $$p$$).

Let $$M$$ be a subgroup of $$G$$ such that $$M/N$$ is a minimal normal subgroup of $$G/N$$. Then, since the order of $$G/N$$ is not a multiple of $$p$$, and $$G/N$$ is solvable, fact (3) yields that $$M/N$$ is an elementary Abelian $$q$$-group for some prime $$q \ne p$$. By fact (5), $$M$$ is itself a normal subgroup of $$G$$.

Let $$Q$$ be a $$q$$-Sylow subgroup of $$M$$ (using fact (4)).

Now, if $$Q$$ is normal in $$G$$, then argue as before replacing $$N$$ by $$Q$$. Note that by our assumption, $$q \in \pi$$, so we land up in case (1).

Let's now consider the case that $$Q$$ is not normal in $$G$$. By Frattini's argument (fact (6)), $$MN_G(Q) = G$$. The product formula yields:

$$\frac{|G|}{|N_G(Q)|} = \frac{|M|}{|M \cap N_G(Q)|}$$

$$Q \le M \cap N_G(Q)$$, so $$|M|/(|M \cap N_G(Q)|)$$ divides $$|M|/|Q|$$, and is hence a power of $$p$$. Thus, $$|G|/|N_G(Q)|$$ is a power of $$p$$. Thus, $$N_G(Q)$$ is a proper subgroup of $$G$$ whose index is a power of $$p$$. By induction, $$N_G(Q)$$ has a $$\pi$$-Hall subgroup, and this must also be a $$\pi$$-Hall subgroup for $$G$$.

Textbook references

 * , Page 200, Exercise 33, Section 6.1 ($$p$$-groups, nilpotent groups and solvable groups)