2-Sylow subgroup is TI implies it is normal or there is exactly one conjugacy class of involutions

Statement
Suppose $$G$$ is a finite group of even order with the property that one (and hence all) of its 2-fact about::Sylow subgroups is a fact about::TI-subgroup. In other words, any two distinct 2-Sylow subgroups intersect trivially. Then, one of the following is true:


 * 1) $$G$$ has exactly one conjugacy class of involutions (i.e., any two elements of order two are conjugate).
 * 2) The 2-Sylow subgroup of $$G$$ is normal, or equivalently, $$G$$ has a unique 2-Sylow subgroup.

Facts used

 * 1) uses::Sylow implies order-dominating
 * 2) uses::Involutions are either conjugate or have an involution centralizing both of them

Proof
Given: A finite group $$G$$ of even order whose 2-Sylow subgroups are TI.

To prove: Either $$G$$ has a single conjugacy class of involutions or $$G$$ has a normal 2-Sylow subgroup.

Proof: Suppose $$G$$ has two distinct conjugacy classes of involutions. Let $$P,Q$$ be any two $$2$$-Sylow subgroups of $$G$$. First, note that every involution is conjugate to some element in any given 2-Sylow subgroup (fact (1), the domination part of Sylow's theorem), so every 2-Sylow subgroup intersects every conjugacy class of involutions. Thus, we may pick $$x \in P, y \in Q$$ as representatives of the two (distinct) conjugacy classes of involutions.

By fact (2), there exists an involution $$z$$ that centralizes both $$x$$ and $$y$$. Since $$z$$ is an involution, it is contained in a 2-Sylow subgroup $$R$$. Because the 2-Sylow subgroups are TI, $$R$$ is the unique 2-Sylow subgroup containing $$z$$. Similarly, $$P$$ is the unique 2-Sylow subgroup containing $$x$$ and $$Q$$

Since $$x$$ and $$z$$ are commuting involutions, $$\langle x,z \rangle$$ is a 2-group. By fact (1), there is a 2-Sylow subgroup containing this. This must equal both $$P$$ and $$R$$, so $$P = R$$. Similarly, $$\langle y,z \rangle$$ is a 2-group, so there is a 2-Sylow subgroup containing it, which must equal both $$Q$$ and $$R$$, so $$Q = R$$. Thus, $$P = Q$$. Since $$P,Q$$ were arbitrary, we conclude that any two 2-Sylow subgroups are equal, so there is a unique 2-Sylow subgroup which must therefore be normal.