Answer to Question #17443 in Abstract Algebra for steve peters

Question

Prove that in any group G, orders of elements ab and ba are equal.

Answer 1

Suppose that ab has order n.

That is, suppose that n is the smallest positive integer such that (ab)^n =e, were e is the identity element of G.

Notice that
(ab)(ab)^(n-1) = (ab)^n = e,
so by uniqueness of inverseses,
(ab)^(n-1) = b^(-1)a^(-1)

Then:
(ba)^n = (ba)(ba) ... (ba)
= b(ab)(ab) ....(ab)a
= b(ab)^(n-1)a
= b( b^(-1)a^(-1) )a
= e

This shows that (ba)^n = e. Furthermore, n is the smallest integer for which this equality holds, since if there were positive m < n such that (ba)^m = e, then a symmetric argument would show that (ab)^m=e, contradicting our definition of n as the order of ab.

Thus n is the order of both elements ab and ba.