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 Thousand Dec 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., $nh=0$.

The following statements need to be fleshed out a bit.

  • 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 $0\leq k<s$, 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 $N$.


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