<s>[INST] Proving Proposition 2.4.3. in Artin I am trying to prove Proposition 2.4.3. in Artin. He leaves this fact unproved. It states:

Let $x$ be an element of finite order $n$ in a group, and let $k$ be an integer that is written as $k = nq + r$ where $q$ and $r$ are integers and $r$ is in the range $0 \leq r < n$. Then:
(a) $x^k = x^r$; (b) $x^k = 1$ if and only if $r = 0$; (c) Let $d$ be the greatest common divisor of $k$ and $n$. The order of $x^k$ is equal to $n/d$.

Here is my attempt. Part (a) is fairly straightforward.

We have $k = nq + r$. $x$ has order $n$, so $x^n = e$, and raising $x$ to any multiple of $n$ also gives $e$ by exponent rules, so we have
\begin{align*}
x^k = x^{nq + r} = x^{nq} x^r = (x^n)^q x^r = e^q x^r = ex^r = x^r.
\end{align*}

I'm only confident on half of part (b).

If $r = 0$, then $k = nq$, so
\begin{align*}
x^k = x^{nq} = (x^n)^q = e^q = e.
\end{align*}
Conversely, suppose $x^k = 1$. Then
\begin{align*}
e = x^k = x^{nq + r} = x^{nq} x^r = (x^n)^q x^r = x^r.
\end{align*}
So $n \mid r$, but $0 \leq r < n$, so this forces $r = 0$.

I am most uncertain on part (c ) and do not know how to start the proof.
 [/INST] You proofs of parts a) and b) are correct.
For part c), note first that $$\left(x^k\right)^{n/d}=\left(x^n\right)^{k/d}=e^{k/d}=e$$ so that the order of $x^k$ divides $\frac nd$.
On the other hand, let $m$ be a positive integer such that $\left(x^k\right)^m=e.$ Then $x^{km}=e$ so that $n\mid km$, and the $$\frac nd \mid\frac kd m\tag1$$  Since $d=\gcd(n,k)$, $\frac nd$ and $\frac kd$ are relatively prime, so $(1)$ implies $$\frac nd\mid m$$
EDIT
In response to OP's comment.  $d$ is the greatest common divisor of $n$ and $k$, so there are positive integers $a$ and $b$ such that $n=ad$, $k=bd$.  Let $c=gcd(a,b)$.  There are positive integers $e,f$ such that $a=ce$, $b=cf$, which gives $n=ecd$, $k=fcd$, so $cd$ is a common divisor of $n$ and $k$.  Since $d$ is the greatest common divisor, we must have $c=1$, which is to say that $\frac nd$ and $\frac kd$ are relatively prime.
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