a group of even order contains an element of order 2


Every group of even order contains an element of order 2.


Let G be a group of even order, and consider the set S={gG:gg-1}. We claim that |S| is even; to see this, let aS, so that aa-1; since (a-1)-1=aa-1, we see that a-1S as well. Thus the elements of S may be exhausted by repeatedly selecting an element and it with its inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, from which it follows that |S| is a multiple (http://planetmath.org/Divisibility) of 2 (i.e., is even). Now, because S(GS)= and S(GS)=G, it must be that |S|+|GS|=|G|, which, because |G| is even, implies that |GS| is also even. The identity elementMathworldPlanetmath e of G is in GS, being its own inverse, so the set GS is nonempty, and consequently must contain at least two distinct elements; that is, there must exist some beGS, and because bS, we have b=b-1, hence b2=1. Thus b is an element of order 2 in G. ∎

Notice that the above is logically equivalent to the assertion that a group of even order has a non-identity element that is its own inverse.

Title a group of even order contains an element of order 2
Canonical name AGroupOfEvenOrderContainsAnElementOfOrder2
Date of creation 2013-03-22 17:11:37
Last modified on 2013-03-22 17:11:37
Owner azdbacks4234 (14155)
Last modified by azdbacks4234 (14155)
Numerical id 20
Author azdbacks4234 (14155)
Entry type Theorem
Classification msc 20A05