Prove: $\operatorname{ord}(a)=\operatorname{ord}(a^{-1})$ We have to look at to cases: 1. $n<\infty$ 2. $n=\infty$ In the first case: let assume $\operatorname{ord}(a)=n$. Therefore: $$e=(a^{-1}a)^n=a^na^{-n}=a^{-n}$$ So $\operatorname{ord}(a^{-1})\leq n$ In the second case: let assume $\operatorname{ord}(a)=\infty$ and $\operatorname{ord}(a^{-1})<\infty$ from the first case we know that $\operatorname{ord}(a)=\operatorname{ord}(a^{-1})<\infty$ contradiction. How can I show the direction $n\leq \operatorname{ord}(a^{-1})$?

Alternatively, we have:

* $\operatorname{ord}(a)=|\langle a \rangle$|

* $\langle a \rangle = \\{ a^n : n \in \mathbb Z \\}$

* $\langle a \rangle = \langle a^{-1} \rangle$




Thus, $\operatorname{ord}(a)=|\langle a \rangle| = |\langle a^{-1} \rangle| = \operatorname{ord}(a^{-1})$.

In words: the two sets are either both infinite or both finite of the same size.