I had a problem with a proof on group and a poster aided me with the proof. But I had problems understanding the final implication relates to GCD and Euclid's lemma:
$$(a^k)^r=e \Rightarrow a^{kr} = e \Rightarrow n | kr \Rightarrow \frac n d | r \cdot \frac k d \Rightarrow \frac n d | r .$$
The last implication is due to the fact that $\gcd \Big( \frac n d, \frac k d \Big) = 1$ and Euclid's lemma.
Thanks in advance.
$\frac{n}{d}\mid r\cdot\frac{k}{d}$ means exactly that any prime that divides $n/d$ also divides $r\cdot k/d$.
Here is where we use Euclid's lemma: if a prime divides the product $r\cdot\frac{k}{d}$, then it must divide one of the factors. But $\gcd(n/d,k/d)=1$, so they don't share any prime factors. Therefore, the prime must divide $r$. This gives $\frac{n}{d}\mid r$.
Hint $\,\ n\mid nr,kr\,\Rightarrow\,n\mid (nr,kr) = (\overbrace{n,k}^{\large d})r\ $ i.e. $\ n\mid dr,\ $ so $\ n/d\mid r $
This proof uses more general gcd properties (e.g. the gcd distributive law) vs. Euclid's Lemma. See here for a few proofs of Euclid's Lemma using these and other properties.
