Prove ord(a) = ord (bab-1). Attempt at the proof. Proof: Let G be a group and let a,b ε G. (=>) Suppose ord(a)=n. That is, an=e. We wish to that the ord(bab-1)=n. If we take an=e and multiply on the left by b and on the right by b-1 then, we have, (1) an=e (2) banb-1=beb-1 (3) banb-1=e. Hence, we have the fact that, (4) an=e=banb-1. Now, if we expand (bab-1)n, we get, (5) bab-1bab-1bab-1..bab-1=banb-1. But, banb-1 = e by our algebra in (1) - (4). Thus, (bab-1)n=banb-1 = e and ord(bab-1)=n. (<=) Let ord(bab-1)=n. That is, (bab-1)n=e. But, as it was shown in (5), (bab-1)n=banb-1 = e. Using the fact that banb-1 = e, we multiply on the left by b-1 and on the right by b and obtain that an=e. Hence, ord(a)=n. QED How does it look? Thanks!!