Prove if $|x|=mn$ and gcd$(a,m)=1$ then $m$ divides $|x^a|$

Question

I don't understand how they got $e^{sn}$ from $x^{a|x^a|sn}$.

This solution is also not something I feel is reasonable to produce without influence. What made them think to multiply $|x^a|n$ of both sides of the GCD equation?

Now, I tried to prove it, and ended up proving that $|x^a|=m$ and was wondering what I did wrong:

Proof.

$(sn)(a) + (t)(mn)=n\Rightarrow\text{gcd}(a,mn)=n$.

$$|x^a| = \frac{|x|}{\text{gcd}(a,|x|)} = \frac{mn}{\text{gcd}(a,mn)} = \frac{mn}{n} = m.$$

Answer 1

$Proof.$ $\ \ (sn)(a) + (t)(mn)=n\Rightarrow (a,mn)=n$.

$|x^a| = \frac{|x|}{(a,|x|)} = \frac{mn}{(a,mn)} = \frac{mn}{n} = m$.

Correct is: $ \frac{mn}{(a,mn)} = m \frac{n}{(a,n)}$ is a multiple of $m$ by $(a,mn) = (a,n)\,$ by $(a,m)=1\,$ and Euclid, and, of course, $\,(a,n)\mid n.$

Answer 2

Since $|x|=mn,$ so $x^{mn}=e$ and thus $$x^{mn|x^a|t}=(x^{mn})^{|x^a|t}=e^{|x^a|t}=e.$$

Also for any element $y,$ we have $y^{|y|}=e.$ In particular, $(x^a)^{|x|^a}=e$ and thus $$x^{a|x|^asn}=\left((x^{a})^{|x^a|}\right)^{sn}=e^{sn}=e.$$

In your attempt, note that $ax+by=c$ only implies $\gcd(a,b)\mid c$ and not the equality. So you cannot conclude $\gcd(a,mn)=n$ but you can conclude that $n=k\gcd(a,mn)$ for some $k$ and so $|x^a|=\frac{m}{k},$ which proves the assertion.