# Involutions are either conjugate or have an involution centralizing both of them

This article gives the statement, and possibly proof, of a statement where the conclusion of the statement involves a disjunction (OR) of two possibilities. The prototypical form is: "every A is a B or a C."

## Statement

Suppose $G$ is a finite group and $x,y$ are two involutions of $G$. Then, at least one of these two is true:

1. $x$ and $y$ are in the same conjugacy class.
2. There exists an involution $z$ centralizing both $x$ and $y$.

## Facts used

1. Dihedral trick

## Proof

Given: A finite group $G$, two involutions $x,y$ of $G$.

To prove: Either $x$ or $y$ are conjugate or there is an involution centralizing both of them.

Proof: By fact (1), $\langle x,y \rangle$ is a dihedral group of order $2m$, where $m$ is the order of $xy$. This dihedral group has $xy$ generating a cyclic subgroup of order $m$ and $x$ acts on this by inverting $xy$.

If $m$ is odd, then $x,y$ are conjugate in $\langle x,y \rangle$, hence in $G$. Otherwise, $m$ is even, in which case the element $z = (xy)^{m/2}$ is an involution centralizing both $x$ and $y$.