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 $\displaystyle a$ has order n. Therefore,ii. Likewise, assume that $\displaystyle a^{-1}$ has order m, with m different from n. Say, $\displaystyle m < n$. Therefore,$\displaystyle a^{-m} = e$

iii. Since $\displaystyle a^n = e$ and $\displaystyle a^{-m} = e$$\displaystyle a^n = a^{-m}$

Multiply both sides by $\displaystyle a^{-m}$ on the right. Then,

$\displaystyle a^n a^{-m} = a^{-m} a^{-m}$

By laws of exponents,

$\displaystyle a^{n-m}= (a^{-m})(a^{-m})$

By the definition of order, on the right we have that

$\displaystyle (a^{-m}) (a^{-m}) = e$

So then we have,

$\displaystyle a^{n-m}= e$

iv. Since $\displaystyle m < n$, as defined at step ii., it follows that$\displaystyle n - m < n$

v. 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 $\displaystyle a^n = e$ holds.

vi. Therefore an element has the same order of its inverse.

Thanks for anyone who helps! (Itwasntme)