Let $x$ be an element in a group $G$. Assume that the order of $x$ is $m$, and also that $x^n = e$. Prove that $m \mid n$.

Question

Let x be an element in a group $G$. Assume that the order of x is m, and also that $x^n = e$ (identity element). Prove that $m \mid n$.

So what I am thinking is that the order of an element is the smallest positive integer $m$ such that $x^m = e$.

There are other integers that satisfy this, but $m$ is the smallest. So $n \geq m$ but also has to be a multiple of $m$. Thus $m \mid n$ in either case.

Is there an intermediate step between this assertion? I more or less just described the idea instead of a formal proof.

Thank you for any and all help.

Answer 1

We combine the fact that $m$ is the order of $x \in G$ with the division algorithm to proceed as follows:

Since $m$ is the order of $x$, it is the smallest positive integer with

$x^m = e; \tag 1$

now use the division algorithm to write

$n = mq + r, \; 0 \le r < m; \tag 2$

then

$x^{mq} = (x^m)^q = e^q = e, \tag 3$

so

$x^r = ex^r = x^{mq}x^r = x^{mq + r} = x^n = e; \tag 4$

if $r \ne 0$, this contradicts the assumption that $m$ is the smallest positive integer such that (1) binds. Thus $r = 0$, $n = mq$, and

$m \mid n. \tag 5$