# A group with an infinite cyclic normal subgroup that has a finite cyclic quotient is abelian

Let $G$ be a group with a normal subgroup $N$ such that $N$ is isomorphic to $\mathbb{Z}.$ Also suppose that $G/N$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for some integer $n \geq 2.$ I need to show that if $n$ is odd, then $G$ is abelian. The following is my attempt:

Because $N$ is infinite cyclic, we have that $N = \langle t \rangle$ for some element $t \in G.$ As $G/N$ is finite cyclic, we have that $G/N = \langle sN \rangle$ for some $s \in G,$ where $s^n \in N.$

Then $G$ is generated by $\{s,t\},$ so to show $G$ is abelian, it suffices to show that the two generators commute.

Consider the element $st \in sN.$ As $N$ is normal, $st \in Ns,$ so $st = t^is$ for some integer $i$. However, I don't know how to proceed any further. In particular, I'm not sure how I am supposed to use the fact that $n$ is odd. The only difference between odd-order cyclic groups and even-order cyclic groups I can think of is that in an odd-order cyclic group, every nontrivial element has an inverse different from itself (which doesn't seem very useful here).

Any help would be much appreciated!

• i.e. solving $0 \to \Bbb Z \to G \to \Bbb Z/n\Bbb Z \to 0$ – Kenny Lau Aug 13 '18 at 7:51
• The important point here is that the automorphism group of ${\mathbb Z}$ has order $2$. – Derek Holt Aug 13 '18 at 7:54

Let $g\in G$, and consider the action $\phi_g$ on $N$. $N$ is normal, so $\phi_g$ is an automorphism. Moreover, $g^n\in N$ implies that $\phi_{g^n}=(\phi_g)^n=\text{id}_N$, because $N$ was abelian. Finally, because $N\cong\mathbb Z$, $\phi_g$ depends exactly on where it sent a generator $t$. If $\phi_g(t)=t^{-1}$ for any $g$, then $t^{(-1)^n}=t$. But $n$ was odd, so that would mean $t^{-1}=t$, a contradiction. So $\phi_g(t)=t$ for each $g$, so in particular $gt=tg$ for each $g\in G$. So each $g$ commutes with all the generators of $N$, so $N\subseteq Z(G)$, so $G/Z(G)\hookrightarrow \mathbb Z/n\mathbb Z$, so it is cyclic. So $G/Z(G)$ is cyclic, so $G$ must be Abelian.