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

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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."


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


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.


Textbook references