Commuting of non-identity elements defines an equivalence relation between prime divisors of the order of a finite CN-group

Statement
Suppose $$G$$ is a finite CN-group (i.e., a finite group that is also a CN-group). Let $$\pi$$ be the set of prime divisors of the order of $$G$$. Define a relation $$\sim$$ on $$\pi$$ as follows: for $$p,q \in \pi$$ (possibly equal, possibly distinct), $$p \sim q$$ if and only if either $$p = q$$ or there exists a $$p$$-Sylow subgroup $$P$$ of $$G$$ and a $$q$$-Sylow subgroup $$Q$$ of $$G$$ such that the following equivalent conditions hold:


 * 1) There exists a non-identity element $$x$$ of $$P$$ and a non-identity element $$y$$ of $$Q$$ such that $$x$$ and $$y$$ commute.
 * 2) Every element of $$P$$ commutes with every element of $$Q$$.

Note that the conditions are equivalent because uses::Sylow subgroups for distinct primes in CN-group centralize each other iff they have non-identity elements that centralize each other. Note also that these conditions are not as strong as the statement that every element of $$p$$-power order in $$G$$ commutes with every element of $$q$$-power order. It is simply a statement about being able to find two specific Sylow subgroups that centralize each other.

The claim is that $$\sim$$ is an equivalence relation on $$\pi$$.

Related facts

 * Sylow subgroups for distinct primes in CN-group centralize each other iff they have non-identity elements that centralize each other is a pre-fact that is necessary to make sense of this.
 * Equivalence classes of primes under commuting equivalence relation on finite CN-group give Hall subgroups that are nilpotent and TI

Facts used

 * 1) uses::Sylow implies order-conjugate
 * 2) uses::Sylow satisfies intermediate subgroup condition

Proof
To prove that a relation is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

Reflexivity
This follows by definition.

Symmetry
This follows by definition.

Transitivity
We want to show that, for $$p,q,r$$ (possibly equal, possibly distinct) primes, $$p \sim q, q \sim r$$ implies $$p \sim r$$. Note that if $$p = r$$, the conclusion follows. If $$p = q$$ or $$q = r$$, it again follows. Thus, we can assume that $$p,q,r$$ are all distinct.

Given: A finite CN-group $$G$$, three distinct primes $$p,q,r$$ such that $$p \sim q, q \sim r$$.

To prove: $$p \sim r$$

Proof: