Given normal subgroup $\cong \mathbb{Z}$ and Quotient $\cong\mathbb{Z}/n\mathbb{Z}$ determine $G$. Let $N$ be a subgroup of $G$ isomorphic to $\mathbb{Z}$. suppose $G/N$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. prove that if $n$ is odd, $G$ is abelian. 
I’ve reduced the problem to showing that if $N$ is central, then $G$ is abelian simply by playing around with the elements. But from there I’m kind of stuck. My main issue is that I cannot find an example of what goes wrong when $n$ is not odd. The only examples for $G$ and $N$ that I can think of are the usual $\mathbb{Z}$ and $n\mathbb{Z}$ respectively, and direct sums like $\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}\times ${1}$ respectively.
Any help is appreciated.
 A: I'll try to put this in a general perspective with group extensions. Suppose we have an exact sequence of groups
$1 \to A \hookrightarrow E \to G \xrightarrow{p} 1$, where $A$ is abelian (Assume wlog that $A \subset E$)
Then we can define a generalized conjugation action of $G$ on $A$ as follows: take an element $g \in G$, lift it to an element $\tilde{g}$ of $E$ (i.e. $p(\tilde{g})=g$) and then for $a \in A$, set $c_g(a):=\tilde{g}a\tilde{g}^{-1} \in A$ (this is an element of $A$ as $A$ is normal.)
There are two things to check:


*

*This is independent of the choice of $\tilde{g}$

*This is actually a group action


1)
Suppose that $\tilde{g}$ and $\tilde{g}'$ are two lifts of $g$, then $x=\tilde{g}^{-1}\tilde{g}' \in A$ by the exactness of the sequence. Thus as $A$ is abelian, for any $a \in A$, we get
$xa=ax$, which means $\tilde{g}^{-1}\tilde{g}'a=a\tilde{g}^{-1}\tilde{g}'$, then if we multiply with $\tilde{g}'^{-1}$ from the right and with $\tilde{g}$ from the left, we get $\tilde{g}a\tilde{g}^{-1}=\tilde{g}'a\tilde{g}'^{-1}$, which shows that the choice of $\tilde{g}$ doesn't matter.
2)
To see that the identity of $G$ acts trivially, note that we can choose as our lift the identity of $E$, since we're free to choose the lift by 1), which of course acts trivially by conjugation on $A$.
If we have $g,h \in G$ and $\tilde{g}$ and $\tilde{h}$ are lifts of $g$ and $h$, then $\tilde{g}\tilde{h}$ is a lift of $gh$, so we get that for $a \in A$, $c_{gh}(a)=\tilde{g}\tilde{h}a(\tilde{g}\tilde{h})^{-1}=\tilde{g}\tilde{h}a\tilde{h}^{-1}\tilde{g}^{-1}=c_g(c_h(a))$, thus this is actually group action.
The idea behind this construction is that in our situation, we can conjugate elements of $A$ with elements of $G$ in a well-defined way, although elements in $G$ are not elements in $E$ in any canonical way (as $G$ is just a quotient, not a subgroup of $E$). How is this related to the question at hand? We have the following result:
$A$ is a central subgroup iff the conjugation action of $G$ on $A$ is trivial.
Proof: it's clear that the action is trivial when $A$ is a central subgroup. Suppose that the action is trivial. Let $e \in E$, then $eae^{-1}=c_{p(e)}(a)$, since by definition $e$ is a lift of $p(e)$, but $c_{p(e)}(a)=a$ by assumption.
This solves the problem, as $\mathrm{Aut}(\Bbb Z)=\Bbb Z/2\Bbb Z$, so if we have an action of $\Bbb Z/n \Bbb Z$, this is a homomorphism $\Bbb Z/n \Bbb Z \to \mathrm{Aut}(\Bbb Z)$ which must be trivial when $n$ is odd.
