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.
I previously posted the question here but the replies were not able to help me (my fault, not theirs).
Question: I have made another attempt at a solution, but unsure if it is correct.
My Attempted Solution:
We start with what the order of an element $x$ means,
$$o(x) = n \implies x^n = e$$
So,
$$ e = x^n = (x^2)^{n/2} = (x^3)^{n/3} = \ldots = (x^m)^{n/m}$$
These exponents $n/m$ are the smallest they can be, because $n$ is already the smallest for $m=1$. To illustrate, $(x^2)^{100 n/2} = e$, but the exponent is 100 times larger than it could be.
However, the exponent $n/m$ is not necessarily an integer. So our task is to find the smallest multiple of $n/m$ which is an integer.
We need to find the smallest $k$ such that $(n/m)k$ that is an integer.
If we think about the unique prime factorisation of $m$ and $n$, then $k$ needs to contain those prime factors of $m$ that are not in $n$. That is $k=m/d$.
This then gives us
$$ e=(x^m)^{kn/m} = (x^m)^{n/d} $$
So $o(x^m) = n/d$.
