Sylow subgroups for distinct primes in CN-group centralize each other iff they have non-identity elements that centralize each other

Statement
Suppose $$G$$ is a finite CN-group: a finite group that is also a  CN-group, i.e., the centralizer of every non-identity element of $$G$$ is nilpotent.

Suppose $$p,q$$ are distinct prime numbers both of which divide the order of $$G$$ (note therefore that this statement does not apply to finite p-groups). Suppose $$P$$ is a $$p$$-Sylow subgroup of $$G$$ and $$Q$$ is a $$q$$-Sylow subgroup of $$G$$. Note that $$P$$ and $$Q$$ are both nontrivial based on the assumption that $$p$$ and $$q$$ both divide the order of $$G$$. Then, the following are equivalent:


 * 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 $$Q$$.

Caveat
Note that these equivalent conditions are not the same as saying that every element of $$p$$-power order commutes with every element of $$q$$-power order. Consider, for instance, the case where $$G$$ is general affine group:GA(1,7), $$p = 2, q = 3$$. Clearly, $$G$$ is a CN-group. It is possible to find a 2-Sylow subgroup and a 3-Sylow subgroup that centralize each other: indeed, take the element of order six in the acting part of the semidirect product and take the 2-Sylow subgroup and the 3-Sylow subgroup inside the subgroup it generates. However, we can also find elements of order 2 and 3 respectively in $$G$$ that do not commute.

In fact, any example where the Sylow subgroups in question are not normal can be used, because the centralizer cannot have a bigger $$p$$-power part than one Sylow subgroup. Explicitly, the centralizer of any non-identity element in the $$q$$-Sylow subgroup, being nilpotent, must have a normal $$p$$-Sylow subgroup, which must hence be precisely $$P$$. In particular, the centralizer of a non-identity element in the $$q$$-Sylow subgroup cannot contain any $$p$$-power order element outside $$P$$. Thus, unless $$P$$ is normal in $$G$$, there are $$p$$-power elements (in fact, all the ones outside $$P$$ work) that do not centralize that particular element of $$Q$$.

Related facts

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

Facts used

 * 1) uses::Equivalence of definitions of finite nilpotent group
 * 2) uses::Prime power order implies not centerless

Direction (2) implies (1)
This is obvious, since both $$P$$ and $$Q$$ are nontrivial.

Direction (1) implies (2)
Given: $$G$$ is a CN-group, $$p,q$$ are primes dividing its order, $$P$$ is a $$p$$-Sylow subgroup of $$G$$, $$Q$$ is a $$q$$-Sylow subgroup of $$G$$. $$x$$ is a non-identity element of $$P$$ and $$y$$ is a non-identity element of $$Q$$. $$x$$ and $$y$$ commute. $$x_2$$ is an arbitrary element of $$P$$ and $$y_2$$ is an arbitrary element of $$Q$$.

To prove: $$x_2$$ and $$y_2$$ commute.

Lemma about commuting being transitive
Given: Elements $$a,b \in P$$, $$c,d \in Q$$, none of them equal to the identity element, such that $$a$$ and $$c$$ commute, $$a$$ and $$b$$ commute, and $$c$$ and $$d$$ commute.

To prove: $$b$$ and $$d$$ commute.

Proof: