Asking for clarification in a group theory proof

Question

The question is:

If $G = \left<a\right>$ have order $rs$, where $\gcd(r,s)=1$, show that there are unique $b,c \in G$ with $b$ of order $r$, $c$ of order $s$, and $a=bc$.

As shown here, the solutions goes as follows:

Since $\gcd(r,s)=1$, $ \exists x,y \in \mathbb{Z}$ such that $xr+ys=1$.
Choose $b=a^{ys}$ and $c=a^{xr}$. It is straightforward to show that $o(b)=r$, $o(c)=s$.

Althouh I can see how it is obvious that $b^r = e$, I cannot understand why $r$ is $o(b)$. Though $(r,s)=1$, $r$ is not prime. Is it wrong to assume that $ \exists i \in \mathbb{Z} $ with $i<r$ such that $iy = r$ so $b^i=a^{iys} = a^{rs} =e$ and $o(b) = i$?

Thanks in advance

Answer 1

1. The order of $H=\langle b,c\rangle$ is on the one hand, upper-bounded [as $H$ is abelian] by $o(b)×o(c)$.

2. The order of $H$ is also, on the other hand, as $a=bc \in H$, exactly $rs$.

3. Then $o(b)|r$ and $o(c)|s$ [because $b^r=c^s=e$].

Then 1.,2., and 3., together imply that $o(b)$ must be exactly $r$ and $o(c)$ must be exactly $s$.

Answer 2

The equation $xr + ys = 1$ implies that $\gcd(x,s) = \gcd(y,r) = 1$, and hence $\gcd(xr,rs) = r$ and $\gcd(ys,rs) = s$.

Therefore $$o(b) = o(a^{ys}) = \dfrac{rs}{\gcd(ys,rs)} = \dfrac{rs}{s} = r$$ and $$o(c) = o(a^{xr}) = \dfrac{rs}{\gcd(xr,rs)} = \dfrac{rs}{r} = s.$$