Your last sentence correctly proves the result.
If G is a finite group of even order, show that there must be an element a not equal to e such that a = a^-1
if I understand correctly, order of a group = number of elements and order of an element is the least integer m such that a^m = e.
so if a does not equal e than a is not the identity element and e^1 =1 so e has order 1
then if a = a^-1 then there is at least one element that is its own inverse, correct?
and by definition, a*a^-1 =e so this is the same as a*a=e or a^2=e ?? so a is of order 2
further, because e is an element that is it's own inverse then there's left an odd number of elements so at least one other element must be it's own inverse.