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

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

Suppose $G$ is a group, $N\trianglelefteq G$ is a normal subgroup, $N$ is cyclic of order $a$, and the quotient group $G/N$ is cyclic of order $b$. Then we can say that $N=\langle n\rangle$ for some $n\in N$ and $G/N=\langle gN\rangle$ for some $gN\in G/N$. Then all of the elements of $N$ are of the form

$$ e,n,n^2,\cdots,n^{a-1} $$

and all of the elements of $G/N$ are of the form

$$N,gN,g^2N,\cdots,g^{b-1}N $$

since $(gN)^i=g^iN$. These cosets partition $G$, so every element $x\in G$ is in a unique $g^iN$, which means it is uniquely expressible as $g^in^j$ for some $0\le i<a$ and $0\le j<b$.

How might a multiplication table for these elements look? We would need to be able to multiply two elements of the form $g^{i_1}n^{i_1}$ and $g^{i_2}n^{i_2}$ and get an element of the form $g^{i_3}n^{i_3}$. In particular, we would need to be able to multiply $(g^0n^1)(g^1n^0)=n\cdot g$ and get $g^rn^s$ for some exponents $r,s$. If we could do that, then we would have a "sliding rule" for simplifying all words involving $g$ and $n$ to the simple form $g^in^j$. For instance, if $ng=gn^3$ then we would have

$$\begin{array}{ll} (gn^2)(gn) & =gnngn \\ & = gn\fbox{$ng$}n \\ & = gn\fbox{$gn^3$}n \\ & = gngn^4 \\ & = g\fbox{$ng$}n^4 \\ & = g\fbox{$gn^3$}n^4 \\ & =g^2n^7. \end{array}$$

The rule $ng=gn^3$ says we can slide $g$ past $n$ to the left, as long as we cube $n$ as we do so. We may do this to move all $g$s to the left. With induction, you can show $ng^k=g^kn^{3k}$ and $n^jg^k=g^kn^{3jk}$ for nonnegative integers $j,k$

The fact that $N\triangleleft G$ is normal means $gng^{-1}\in N$ so $gng^{-1}=n^c$ for some $0\le c<a$, and also $g^{-1}ng=n^d$ for some $0\le d<a$. Note that $\varphi(x)=gxg^{-1}$ is an automorphism of $G$ (i.e. $\varphi(xy)=\varphi(x)\varphi(y)$), and the fact that $N$ is normal means $\varphi(N)=N$, so $\varphi$ restricts to an automorphism of $N$. Every element of $N$ looks like $n^k$, and $\varphi(n^k)=\varphi(n)^k=(n^c)^k=(n^k)^c$, or in other words $\varphi(x)=x^c$ for all $x\in N$. In particular $n=g(g^{-1}ng)g^{-1}=\varphi(n^d)=n^{cd}$ implies that $cd\equiv 1$ mod $a$, so $c,d$ are units mod $a$ (i.e. they are coprime to $a$).

Going back to the multiplication table idea, the relations $gng^{-1}=n^c$ and $g^{-1}ng=n^d$ give us the sliding rules $ng^{-1}=g^{-1}n^c$ and $ng=gn^d$. This allows us to move both $g$ and $g^{-1}$ to the left of $n$ (and hence powers of $n$ by induction).

$G$ is what is called a semidirect product $N\rtimes Q$ (where $Q\cong G/N$). In general, this means $G$ has two subgroups $N$ and $Q$, with $N$ normal, such that every element is expressible as $nq$ uniquely, and conjugating elements of $N$ by an element of $Q$ is an automorphism of $N$, hence the tacit existence of a map $Q\to\mathrm{Aut}(N)$. In this case, we get a group with presentation

$$ G=\langle n\rangle\rtimes\langle g\rangle=\langle n,g\mid n^a=e,g^b=e,gng^{-1}=n^c \rangle $$

with $0\le c<a$ coprime to $a$. What this means is that every element of $G$ is expressible using just $n$ and $g$, and every true equation involving $g$ and $n$ can be deduced using just the three relations.

Note that the phrase "generated by at most two elements" means there exists a generating set having two or less elements. (We've proven this: $\{g,n\}$ is a generating set.) It does not mean it is impossible to generate the group with more than two elements. In fact, this should be obviously false: if $S\subseteq G$ is any generating set for a group $G$, and $S\subseteq A\subseteq G$ for any other subset $A$, then $A$ is also a generating set. In particular, $G$ itself will be a generating set for itself.

$G$ itself may not be cyclic or abelian. Indeed if $c\ne 1$, then $gng^{-1}=n^c\ne n$, so $g$ and $n$ do not commute. However, it is possible for $G$ to be abelian but not cyclic, and it is possible for $G$ to be cyclic. If $c=1$ then $g$ and $n$ commute, which implies $G$ is commutative and $G\cong\mathbb{Z}/a\mathbb{Z}\times\mathbb{Z}/b$. If $a$ and $b$ are not coprime, this is not cyclic. But if $a$ and $b$ are coprime, then $ \mathbb{Z}/a\mathbb{Z}\times\mathbb{Z}/b\mathbb{Z}\cong\mathbb{Z}/ab\mathbb{Z}$ is cyclic.

You may be interested in:

• The Chinese Remainder Theorem.
• The Classification of Finitely Generated Abelian Groups

Let's say you want to go above and beyond the call of duty, though. Call a generating set $S$ minimal if it has no proper subset which is also a generating set. Does $G$ have a minimal generating set with more than two elements? Yes, it can. Let's consider the simplest case: when $G$ is cyclic. For instance, suppose it is cyclic of order $pqr$ where $p,q,r$ are distinct primes. Then

$$G\cong\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/q\mathbb{Z}\times\mathbb{Z}/r\mathbb{Z}$$

and the latter has a minimal generating set of size three, namely

$$\{(1,0,0),(0,1,0),(0,0,1)\}. $$

The Wikipedia page on the Chinese Remainder Theorem explains how to get the corresponding three elements of $\mathbb{Z}/pqr\mathbb{Z}$ if you're so inclined. Let $\overline{p},\overline{q},\overline{r}$ be multiplicative inverses of $p,q,r$ mod $qr,rp,pq$ respectively. Then the generating set is $\{\overline{p}qr,p\overline{q}r,pq\overline{r}\}$.

So far we've labelled these groups $G$ with ordered triples $(a,b,c)$ where $0\le c<a$ is coprime to $a$. (This allows $a$ and/or $b$ to be infinite. If $a=\infty$ then $c=\pm1$.) This is probably a redundant listing, i.e. there could be different ordered triples that label isomorphic groups. Finding an irredundant enumeration sounds like an interesting problem that I'll shelve for now.

In general, whenever we have a bunch of inclusions

$$ G_0\subset G_1\subset \cdots \subset G_\ell $$

where each $G_i$ is normal in $G_{i+1}$, we can think of the successive quotients

$$G_1/G_0, ~ G_2/G_1, ~ \cdots, ~ G_{\ell}/G_{\ell-1}$$

as "pieces" of the group. If we make the chain as long as possible (so the chain cannot be made longer by inserting any new group strictly between existing ones) it is called a composition series, and the quotients are called composition factors. The Jordan-Holder theorem says that the multiset (set with multiplicity) of composition factors (up to isomorphism) is uniquely determined by the group. In some sense, composition factors are to groups as atoms are to molecules, or prime numbers are to natural numbers. When every composition factor is abelian (as in this case, $N$ and $G/N$ are abelian), the group is called solvable.

If $P$ is a property that a group can have, then we say $G$ is meta-$P$ if it has a normal subgroup $N$ such that $N$ and $G/N$ have the property $P$. This is why our $G$'s are called metacyclic. More generally, if $P$ and $Q$ are group properties, then $G$ is $P$-by-$Q$ if it has a normal subgroup $N$ such that $N$ has property $P$ and $G/N$ has property $Q$.

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.