Prove that in any group, an element has the same order of its inverse
Am I on the right track?
This will be a proof by contradiction.
i. Assume has order n. Therefore,ii. Likewise, assume that has order m, with m different from n. Say, . Therefore,iii. Since and
iv. Since , as defined at step ii., it follows thatv. But the result of iv. contradicts the claim that a has order n, since by definition this means that n is the least positive integer such that holds.
Multiply both sides by
on the right. Then,
By laws of exponents,
By the definition of order, on the right we have that
So then we have,
vi. Therefore an element has the same order of its inverse.
Thanks for anyone who helps! (Itwasntme)