Normal Hall implies permutably complemented

Name
This result is sometimes called Schur's theorem, and is considered a part of the Schur-Zassenhaus theorem, which also asserts that any two permutable complements to a normal Hall subgroup are conjugate. The second half is available at Hall retract implies order-conjugate.

Verbal statement
Any normal Hall subgroup of a group is permutably complemented.

Statement with symbols
Suppose $$N$$ is a normal Hall subgroup of a group $$G$$. Then there exists a subgroup $$H \le G$$ such that $$N \cap H$$ is trivial and $$NH = G$$. Note that this makes $$H$$ a Hall subgroup as well.

Equivalently, if $$N$$ is a normal $$\pi$$-Hall subgroup of $$G$$, then $$G$$ possesses a $$\pi'$$-Hall subgroup: a Hall subgroup corresponding to the primes not in $$\pi$$.

Facts used

 * 1) uses::Abelian normal Hall implies permutably complemented: This was the original version of the theorem proved by Schur.
 * 2) uses::Sylow subgroups exist
 * 3) uses::Frattini's argument
 * 4) uses::Normal Hall satisfies transfer condition
 * 5) uses::Equivalence of definitions of finite nilpotent group
 * 6) uses::Center of normal implies normal

Proof
We prove this claim by induction on the order of $$G$$. Rather, we reduce the claim to (1) by induction on the order of $$G$$.

Given: A group $$G$$, a normal Hall subgroup $$N$$ of $$G$$.

To prove: $$N$$ has a permutable complement in $$G$$.

Proof: Note that for any prime $$p$$ dividing the order of $$N$$, a $$p$$-Sylow subgroup of $$N$$ is also $$p$$-Sylow in $$G$$. Note also that by fact (2), $$p$$-Sylow subgroups exist for all primes $$p$$.


 * 1) Case where there is a prime $$p$$ dividing the order of $$N$$, such that a $$p$$-Sylow subgroup $$P$$ of $$N$$ is not normal in $$G$$: Let $$M = N_G(P)$$. $$NM = G$$ by fact (3) (Frattini's argument). Further, by fact (4), $$N \cap M$$ is a normal Hall subgroup of $$M$$, so by induction, there exists a subgroup $$H$$ of $$M$$ that is a permutable complement in $$M$$ to $$N \cap M$$. We claim that $$H$$ is a permutable complement to $$N$$ in $$G$$:
 * 2) * $$H \cap N$$ is trivial: Since $$H \le M$$ and $$H \cap (N \cap M)$$ is trivial, we obtain that $$H \cap N$$ is trivial.
 * 3) * $$HN = G$$: We have $$M = H(N \cap M) \le HN$$. Also, $$N \le HN$$. Thus, $$NM \le HN$$. But as observed earlier, by fact (3), $$NM = G$$. this forces $$HN = G$$.
 * 4) Case where $$N$$ is non-abelian and, for every prime $$p$$ dividing the order of $$N$$, a $$p$$-Sylow subgroup of $$N$$ is normal in $$G$$: In this case, each of the $$p$$-Sylow subgroups of $$N$$ is normal in $$N$$, hence $$N$$ is a direct product of its Sylow subgroups. Thus, $$N$$ is nilpotent. In particular, the center $$Z(N)$$ is nontrivial. By fact (6), $$Z(N)$$ is normal in $$G$$. Consider the group $$G_1 = G/Z(N)$$. $$N_1 = N/Z(N)$$ is a normal Hall subgroup of $$G_1$$, so by induction, there exists a permutable complement $$H_1$$ to $$N_1$$ in $$G_1$$. Suppose $$K$$ is the inverse image of $$H_1$$ under the quotient map $$G \to G/Z(N)$$. Note that $$K$$ is a proper subgroup of $$G$$, since $$Z(N)$$ is a proper subgroup of $$N$$ by the assumption that $$N$$ is not abelian.$$K/Z(N)$$ and $$Z(N)$$ have relatively prime orders, so $$Z(N)$$ is a normal Hall subgroup of $$K$$. Thus, there exists a permutable complement $$H$$ to $$Z(N)$$ in $$K$$.
 * 5) Case that $$N$$ is abelian: This is covered by fact (1).