Nilpotent Hall subgroups of same order are conjugate

History
The result was proved by Wielandt in 1954.

Verbal statement
In a finite group, any two nilpotent Hall subgroups of the same order are conjugate subgroups.

Statement with symbols
Suppose $$G$$ is a finite group and $$\pi$$ is a set of primes. Let $$H$$ and $$K$$ be Hall $$\pi$$-subgroups of $$G$$. Then, $$H$$ and $$K$$ are conjugate subgroups inside $$G$$: in other words, there exists $$g \in G$$ such that $$gHg^{-1} = K$$.

Facts used

 * 1) uses::Equivalence of definitions of finite nilpotent group: Specifically, the fact that in a finite nilpotent group, all the Sylow subgroups are normal
 * 2) uses::Normality is upper join-closed: If $$N \le H, L \le G$$ and $$N$$ is normal in both $$H$$ and $$L$$, then $$N$$ is normal in $$G$$
 * 3) uses::Sylow implies order-conjugate: Specifically, the fact that any two $$p$$-Sylow subgroups of a finite group are conjugate

Proof
Given: A finite group $$G$$, a set $$\pi = \{ p_1,p_2, \dots, p_r\}$$ of prime divisors of $$G$$. Two nilpotent Hall subgroups $$H,K$$ of $$G$$

To prove: $$H$$ and $$K$$ are conjugate.

Proof: We prove the result by a double induction: induction on the order of $$G$$, and, for a given $$G$$, induction on the size of $$\pi$$.

Outer induction on order: Let's assume the result is true for all groups of order smaller than the order of $$G$$.

Inner induction on size of set of primes: Note that the result is true if $$\pi$$ has size one, by Fact (3). The hard part is the inductive step. We will assume that the result holds for the particular group $$G$$ for all smaller sized sets of primes.

Textbook references

 * , Page 236, Exercise 2 (Chapter 6)