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Let $G$ be a group and let $x \in G$. Prove that $x$ and $x^{-1}$ have the same order.

Solution: Recall that the order of a group element is either a positive integer or infinity.

Suppose $|x|$ is infinite and that $|x^{-1}| = n$ for some $n$. Then $$x^n = x^{(-1) \cdot n \cdot (-1)} = ((x^{-1})^n)^{-1} = 1^{-1} = 1,$$ a contradiction. So if $|x|$ is infinite, $|x^{-1}|$ must also be infinite. Likewise, if $|x^{-1}|$ is infinite, then $|(x^{-1})^{-1}| = |x|$ is also infinite.

Suppose now that $|x| = n$ and $|x^{-1}| = m$ are both finite. Then we have $$(x^{-1})^n = (x^n)^{-1} = 1^{-1} = 1,$$ so that $m \leq n$. Likewise, $n \leq m$. Hence $m = n$ and $x$ and $x^{-1}$ have the same order.