Nature of $G$ when $N$ is cyclic, normal subgroup of $G$ and $G/N$ is cyclic

Question

Let $N$ be a normal subgroup of $G$ and both groups $N$ and $G/N$ are cyclic. I need to prove that $G$ is generated by at most two elements.

To that effect, what sorts of things do we know about $G$ if $N$ is a subgroup of $G$ and $N$ and $G/N$ are both cyclic?

I know that all cyclic groups are abelian, so $N$ and $G/N$ must both be abelian as well. Does that necessarily mean that $G$ itself is abelian? Or cyclic?

Really, I am having a lot of trouble proving that $G$ is generated by at most two elements, especially in the case where $G$ is generated by a single element (i.e., is cyclic) and showing that it is not possible for $G$ to be generated by more than two elements.

Could somebody please provide me with insight into this?

Thank you.

Answer 1

No. These groups are called metacyclic. A semidirect product of cyclic groups is metacyclic, but is not necessarily cyclic. The simplest examples are the dihedral groups.

Answer 2

For the generated by two elements part: $N$ is cyclic, so its generated by a single element, call it $x$, with $|x|=n$ for some positive natural number. Then, $G/N$ is cyclic, so it is generated by a single element, call it $y$.
Thus, all the cosets look like $y^kN$. Now, every element $g\in G$ is a member of exactly one coset, so $\exists k$ such that $g\in y^kN$. But to be in the coset, that means $\exists n\in N$ such that $g=y^kn$. But $n\in N$, so $\exists j$ such that $n=x^j$. Thus $g=y^kx^j$, so every element of $G$ can be written as a product of $y$ and $x$, and thus is generated by those two elements

Answer 3

Let $p:G\rightarrow G/N$ the projection map, $p([G,G])=1$ where $[G,G]$ is the commutator of $G$, thus $[G,G]\subset N$ and is nilpotent. $G$ is a solvable group of length $\leq 2$.

Answer 4

As I commented these groups are called 'metacyclic' and Hempel C. E. showed in his 2000 paper 'Metacyclic groups' the next characterization theorem about them:

A group G is metacyclic with a kernel of order $\gamma$ and of index $\alpha$ if and only if it has a presentation $G=\langle a,b|a^\alpha=b^\beta,b^\gamma=1,a^\alpha=b^\delta\rangle$ where $\alpha$, $\beta$, $\gamma$, $\delta$ are positive integers such that $\gamma|\delta$, $\alpha−1$ and $\gamma|\beta(\delta−1)$.