Proving Even Order Group G Has Element a=/e Satisfying a^2=e

[HallsofIvy]

I'm a bit confused by the wording of this statement. Do you mean that since we now have |G|=odd, and then we multiply that number of sets with (2), then the order will actually become even?

If I understand what you are saying here, yes, that's the basic idea. IF G contains no member a, other than the identity e, such that a^-1= a, then we can pair all non-identity members, {a, a^-1}. If we multiply the number of such sets by 2, we get the number of non-identity members which must be even. Since there is one identity, e, the total number of members of G, |G|, must be odd.

Here's an example. Suppose we have a group with 4 members, e, x, y and z where e is the group identity. Suppose further that there is no member, other than e, such that a^-1= a. e is its own inverse. What is the inverse of x? Since x is not its own inverse, it must be either y or z. Suppose it is y: we can pair {x,y}. Now that leaves z. z is not, by hypothesis, its own inverse, but there is nothing else left to pair it with!

It occurs to me that this requires we know that the inverse of z cannot be either x or y because they are "already paired"- theorem: if a^-1= b and c^-1= b, the a= c. Do you see why that is true? Suppose a^-1= b and c^-1= b. Look at a(cb) and (ac)b.