proof Let and , each of prime order , be subgroups of a group . If , prove .
If and are of prime order , every element and have order or order by the corollary to Lagrange's Theorem. However, only the identity can have order and the identity is unique. Thus both and have elements of order . Since the orders of those elements are in a group of order , both and have elements which generate their respective groups. Now, since and any , , generates exactly , . Likewise for any , , . Since and are both groups, they both must contain the identity . Therefore, if , where and are of prime order , it must be true that .