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Let $G$ be a group with a normal subgroup $N$ such that $N$ is isomorphic to $mathbbZ.$ Also suppose that $G/N$ is isomorphic to $mathbbZ/nmathbbZ$ 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!

Let $gin G$, and consider the action $phi_g$ on $N$. $N$ is normal, so $phi_g$ is an automorphism. Moreover, $g^nin N$ implies that $phi_g^n=(phi_g)^n=textid_N$, because $N$ was abelian. Finally, because $Ncongmathbb Z$, $phi_g$ 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 $gin G$. So each $g$ commutes with all the generators of $N$, so $Nsubseteq Z(G)$, so $G/Z(G)hookrightarrow mathbb Z/nmathbb Z$, so it is cyclic. So $G/Z(G)$ is cyclic, so $G$ must be Abelian.