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## The powers of a group element of infinite order are pairwise distinct

Let $G$ be a group, and let $x \in G$ be an element of infinite order. Prove that the elements $x^k$ with $k \in \mathbb{Z}$ are all distinct.

Solution: Suppose to the contrary that $x^a = x^b$ for some (distinct) integers $a$ and $b$; without loss of generality say $a < b$. Then we have $x^{b-a} = 1$ and $0 < b-a$. This is a contradiction since $x$ has infinite order; thus no such $a$ and $b$ exist.