So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.

Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because

. We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element

is not in A we know that

.

Thus there is at least one element a in G that is not the identity that has the property

. (End of proof.)

Wow. That took a lot more writing than I thought it would.

Thanks for the tip!

-Dan