**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.23**

Let $G$ be a group. Suppose $x \in G$ with $|x| = n < \infty$. If $n = st$ for some positive integers $s$ and $t$, prove that $|x^s| = t$.

Solution: We have $$(x^s)^t = x^{st} = x^n = 1,$$ so that $|x^s| \leq t$.

Now suppose that in fact $|x^s| = u < t$ for some positive integer $u$. Then $$(x^s)^u = x^{su} = 1,$$ where $su < st = n$. This gives a contradiction, so that no such $u$ exists. Hence $|x^s| = t$.