Algebra Solutions: Characterization of the order of powers of a group element

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 $|x^s| = t$ .

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 $|x^s| = t$ .

Algebra Solutions

Characterization of the order of powers of a group element

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