Question

Prove that order of a is equal to order of inverse of a

Answer 1

Suppose that aa has infinite order. We show that a−1a−1 cannot have finite order. Suppose to the contrary that (a−1)m=e(a−1)m=e for some positive integer mm. We have by repeated application of associativity that

am(a−1)m=e.am(a−1)m=e.

It follows that am=eam=e.

Answer 2

Let anan be ee, then e=(aa−1)n=an(a−1)n=e(a−1)n=(a−1)ne=(aa−1)n=an(a−1)n=e(a−1)n=(a−1)n.

Let (a−1)n=e(a−1)n=e, then e=(aa−1)n=an(a−1)n=ane=ane=(aa−1)n=an(a−1)n=ane=an.

So, an=e⟺(a−1)n=ean=e⟺(a−1)n=e.

$hope \: helps$