Problem Statement: Prove that a group of even order must have an element of order .
Proof: Let be a group such that and consider the set . is the set of all elements of that are not of order .
I claim that is even. Let , then and so as well. So no element may be in unless it’s inverse is also an element of , thus, is even. Let .
Note that any element of is either in or in , by construction. Furthermore, and and so it follows that . This implies that and so it must be that is even. Let .
where . Since must be it follows that there is always some that is not the identity element. Since it follows that which implies that and so .
Thus, any group of even order must have at least one element of order .
Reflection: The key to this proof was breaking the group into two subsets, elements of order and elements that were not of order . The other major idea is that every element is the inverse of it’s inverse. i.e. . This forces our set to have even order which in turn forces the set we are interested in to also have even order. So, even though we know the identity is always in , since the order is even there must always be at least one other, non-identity element in as well.