Exercise 1.1.20 ($x$ and $x^{-1}$ have the same order)

Proof. By Exercise 19, ${(x^{-1})}^n={(x^n)}^{-1}=x^{-n}$ for any $n\in \mathbb{Z}$. If $x^{-1}$ has infinite order, then ${(x^{-1})}^n\ne 1$ for all integers $n$, thus $x^{-n}\ne 1$ for all $n$, so $x$ has infinite order. Similarly the converse is true, so

Suppose now that $x$ has finite order. For all $n\in \mathbb{Z}$,

So for all $n\in \mathbb{Z}$

For $n=|x|$, we obtain that $|x|$ divides $|x^{-1}|$, and for $n=|x^{-1}|$, this shows that $|x^{-1}|$ divides $|x|$. Therefore

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