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This problem is No. 2.21 from the book "Introduction to the theory of groups," by Rotman.

Let $G = \langle a \rangle$ be a cyclic group of order $st$, where $\gcd (s,t) = 1$. Show that there are unique $b, c \in G$ with $b$ of order $s$ and $c$ of order $t$, and $a=bc$.

I have tried proving there's always a unique integer $y$ where $ys\equiv 0\pmod {st}$ and $(y+1)t\equiv 0\pmod {st}$, and then take $a^{y+1}$ and $a^{-y}$ as my generators after proving non of them generates a smaller group than $|s|$ and $|t|$ respectively, but failed to do so.

Hints\solutions are welcome.

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  • $\begingroup$ I am sorry, i miswrote a as x. $\endgroup$ – Fuseques Jan 24 at 10:00
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    $\begingroup$ in my opinion, since $s,t$ are coprime, there are $\zeta,\xi\in\mathbb{Z}$ such that $\xi s+\zeta t=1$, then $a=a^1=a^{\xi s+\zeta t}$, and continue from here $\endgroup$ – Alessandro Jan 24 at 10:07
  • $\begingroup$ yes sorry again. about ζ and ξ: in order for them to generate groups of order $|s|$ and $|t|$ they need to be coprimes to them respectively, i kind of got stuck there. $\endgroup$ – Fuseques Jan 24 at 10:52
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    $\begingroup$ Chinese remainder theorem says that such integers do exist. Looking at the difference of two such integers says that they must be at least $st$ apart. $\endgroup$ – Ivan Neretin Jan 24 at 11:43
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Since $gcd(s,t)=1$, there exist integers $n$ ad $m$ such that $ns+mt=1$. In fact, these integers can be chosen so that $\lvert n \rvert <t$ and $\lvert m \rvert <s$. Now, choose $b=a^{mt}$ and $c=a^{ns}$.

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  • $\begingroup$ Thanks. While I see why $n$ and $m$ are unique $\pmod{st}$, and while $a^{mt}$ cannot generate a group with order bigger than $s$, why is it the case that it can never generate a group of order smaller than $s$? $\endgroup$ – Fuseques Jan 25 at 6:40
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    $\begingroup$ Oh sorry, that is not clear. It cannot because $gcd(m,s)=1$ (Otherwise divide everything by $gcd(m,s)$ in the equation $ns+mt=1$ to get a contradiction). $\endgroup$ – Uğur Cin Jan 25 at 6:58

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