# Finding unique cyclic subgroup generators.

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.

• I am sorry, i miswrote a as x. – Fuseques Jan 24 at 10:00
• 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 – Alessandro Jan 24 at 10:07
• 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. – Fuseques Jan 24 at 10:52
• 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. – Ivan Neretin Jan 24 at 11:43

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 $$\lvert n \rvert and $$\lvert m \rvert . Now, choose $$b=a^{mt}$$ and $$c=a^{ns}$$.
• 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 $s$? – Fuseques Jan 25 at 6:40
• 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). – Uğur Cin Jan 25 at 6:58