The GCD of any two orders is the same order

Proposition

Let $G$ be an arbitrary Group, $x \in G$ and let $m, n \in Z$ . If $x^{n} = x^{m} = 1$ then $x^{d} = 1$ where $d = gcd (m, n)$ . In particular if $x^{m} = 1$ for some $m \in Z$ (ie: $| x | = m$ ) then $| x |$ divides $m$ .

Proof

By Lecture 3 - Divisibility, Euclid's Algorithms#Euclid's Division Algorithm, then $\exists r, s$ where $d = m r + n s$ where $d = gcd (m, n)$ . Thus:

x^{d} = x^{m r + n s} = (x^{m})^{r} (x^{n})^{s} = 1^{r} 1^{s} = 1

This proves the first assertion.

If $x^{m} = 1$ then let $n = | x |$ . If $m = 0$ then certainly $n | m$ so assume $m \neq 0$ . Since some nonzero power of $x$ is the identity, $n < \infty$ . Let $d = gcd (m, n)$ so by the preceding result then $x^{d} = 1$ . Since $0 < d \leq n$ and $n$ is the smallest positive power of $x$ which gives the identity, we must have $d = n$ , that is, $n | m$ as desired.

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