The following is an exercise from "Guide to Abstract Algebra", 1st edition, Carol Whitehead.
Let $G$ be a group and $x \in G$. Suppose $o(x)= n$ and $m$ is an integer such that $\gcd(m, n) = d$. Prove that $o(x^m) = n/d$.
Here $o(x)$ is the order of the element $x$, the smallest positive integer $r$ such that $x^r = e$, the identity element.
Question: I am unsure of my solution. Is it correct?
My Solution Attempt:
First we write what the order means,
$$o(x) = n \implies x^n = e$$
Then,
$$ (x^n)^k = (e)^k = e $$
where $k$ is a positive integer.
Now we want to re-arrange this in terms of $x^m$, so we can write $k=mq$.
$$e = (x^n)^{mq} = (x^m)^{nq}$$
Now, $nq$ is the order of element $(x^m)$ if $nq$ is the smallest positive integer. This is the case when $nq=n/d$, because $d$ is the largest integer that still allows $n/d$ to remain an integer.
Hence $o(x^m) = n/d$.
My Thoughts:
$e=(x^m)^{nq}$ is clearly true, but in seeking to minimise the exponent ${nq}$ I can't justify why we have to retain $n$. If it can be justified, then the next step of choosing $q=1/d$ is justifiable.
This is false, if you let $nq=n/d$, it gives
$$q=\frac{1}d$$
But in general, $m, n$ are not co-primes, hence $d=\gcd(m, n)\ge 2$, but $q\in\mathbb{N}^+$, which means
$$q\ge1>\frac{1}d$$
Contradiction.