# Finite abelian group $G$, with $N \triangleleft G$ cyclic and $G/N$ cyclic [closed]

Let $$G$$ be a finite abelian group and $$N \triangleleft G$$. If both $$N$$ and $$G/N$$ are cyclic and $$GCD(\vert N \vert,\vert G/N \vert)=1$$ then $$G$$ is cyclic.

I don't know how to prove that.

## closed as off-topic by Derek Holt, Trevor Gunn, Lord_Farin, Namaste, Don ThousandDec 18 '18 at 19:36

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Suppose that $$g$$ generates $$N$$ and $$h+N$$ generates $$G/N$$. Let $$|N|=s$$ and $$|G/N|=t$$, then $$|G|=st$$. Suppose that $$o(h)=n$$, i.e., $$nh=0$$.
• We may conclude that $$n$$ is a multiple of $$t$$. In particular, $$0+N=nh+N=n(h+N)$$, so $$n$$ is a multiple of the order of $$h$$, i.e., $$t$$. Therefore, $$n=mt$$ for some $$m$$.
• Consider $$th$$. Since $$th+N=t(h+N)=0+N$$, we may conclude that $$th\in N$$. If $$th$$ is a generator for $$N$$, then we may conclude that $$h$$ generates $$G$$: since $$0=nh=m(th)$$ and $$th$$ is a generator for $$N$$, it must be that $$m$$ is a multiple of $$s$$, so $$st\mid n$$. Then, by Lagrange, we may conclude that $$st=n$$.
• Now, suppose that $$th$$ is not a generator for $$N$$. Then $$th=kg$$ for some $$0\leq k, where $$k$$ is not relatively prime to $$s$$. Our goal is to find an $$a$$ so that $$kg=t(ag)$$. In other words, $$k-ta$$ is divisible by $$s$$. Rewriting this, we have $$k-ta\equiv 0\pmod{s}$$. Since $$t$$ is relatively prime to $$s$$, $$t$$ is invertible modulo $$s$$, so $$a\equiv t^{-1}k\pmod s$$. Choose any such representative for $$a$$.
• In this case, $$h+(1-a)g$$ is a generator for $$N$$. Namely, $$t$$ must divide the order of $$h+(1-a)g$$ by the argument above, but $$t(h+(1-a)g)=th+tg-tag=tg$$. Since $$t$$ is relatively prime to $$s$$, by the theory of cyclic groups, $$tg$$ is a generator for $$N$$.