Ex $2.21(i):$ The question as verbatim.
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 $a = bc$.
$\mathbf{NB:}$ I can't seem to get $a = bc$, unless there's an errata in the question!?
To show uniqueness, let $e = b_{1}^{r} = b_{2}^{r}$, then by Lagrange's there is a subgroup $<b_{1}> = <a^{s_{1}}>$ and $<b_{2}> =<a^{s_{2}}>$, then $e = (a^{s_{1}})^{r} = (a^{s_{2}})^{r} \Rightarrow s_{1}r = s_{2}r \Rightarrow s_{1} = s_{2}$, hence uniqueness is obtained, same argument can be applied for the element $c\in G$.
So, $<b> = <a^{s}>$ and $<c> = <a^{r}>$, and since $gcd(r,s)=1$, then there are integers $k,l$ such that $1 = rk + sl$, hence $a = a^{rk + sl} = (a^{r})^{k}(a^{s})^{l} = b^{k}c^{l}$.
Could you kindly guide are to where my pitfall is in my argument! It'll be much appreciated.
The question is asking to show that there are unique $b, c$ such that $o(b) = r$, $o(c) = s$, and $a = bc$.
Since $\gcd(r, s) = 1$, there exist $x, y \in \mathbb{Z}$ such that $xr + ys = 1$.
Choose $b = a^{ys}$ and $c = a^{xr}$. It is straightforward to show that $o(b) = r$, $o(c) = s$.
One has $$bc = a^{ys} \cdot a^{xr} = a^{xr + ys} = a$$ as required.
To prove uniqueness, assume that $a = b_1c_1 = b_2c_2$ where $o(b_1) = o(b_2) = r$ and $o(c_1) = o(c_2) = s$. Rearranging gives $b_2^{-1}b_1 = c_2c_1^{-1}$, and so $(c_2c_1^{-1})^r = 1$. Powering to $x$ gives $c_2c_1^{-1} = 1$, and so $c_1 = c_2$. It follows immediately that $b_1 = b_2$.