Order of an element and its inverse in a group

Question

Let order of element $a$ be $m$, then prove that order of $a'$ ( inverse of $a$) is also $m$ in a group. For proving this I used the fact- Inverse of inverse element is again the same element. $(a')'=a$, then taking the power $-m$ to both sides for m in set of positive integers, we get $((a')')^{-m} =a^{-m}$ , implies that $(a')^m=(a^m)^{-1}$, implies that = $e^{-1}$ = $e$ we get the proof finally. I think the proof is not systematic. Need systematic proof.

Answer 1

Just prove that $(a')^m = (a^m)'$. Since $e$ is the only element whose inverse is $e$, this implies that the powers of $a$ which equal $e$ are the same as the powers of $a'$ which equal $e'=e$.

Answer 2

Heres the answer to my question:- Let n be the order of the element a and m be the order of its inverse a^-1. Claim: n=m We have O(a)=n

implying that a^n=e

implying that (a^n)^-1=e^-1

implying that (a^-1)^n=e

implying that O(a^-1) is less than or equal to n.

implying that

m is less than or equal to n...........(1) Also O(a^-1)=m implies that (a^-1)^m= e

implies that (a^m)^-1= e implies that a^m =e implies that O(a) is less than or equal to m implies that n is less than or equal to m........(2).

From (1) & (2), we get m=n.

Answer 3

Here is a slightly different take.

We have $\langle a \rangle = \{ a^n : n \in \mathbb Z \} = \{ a^{-n} : n \in \mathbb Z \} = \{ (a^{-1})^n : n \in \mathbb Z \} = \langle a^{-1} \rangle$.

Therefore, $o(a) = |\langle a \rangle| = |\langle a^{-1} \rangle| = o(a^{-1})$.