# 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-Sylow subgroup (?)s is a 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.

## 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 $Q$ is the unique 2-Sylow subgroup containing $y$.

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.