Show that the factor group $G/N$ is cyclic of order 4. 
Let $ G $ be the set of all matrices in $GL_2(\mathbb{Z}_5)$ of the form
  $ \begin{bmatrix}
            m & b\\
            0 & 1
        \end{bmatrix} $.
  And $N$ is the set of matrices in $G$ of the form $ \begin{bmatrix}
            1 & c\\
            0 & 1
        \end{bmatrix} $ with $c\in \mathbb{Z}_5$.
  Show that the factor group $ G/N $ is cyclic of order 4.

I have already shown that $N$ is a normal subgroup of $G$. But I do not understand what it means for $G/N$ to be cyclic or how to show it. Does that mean that for some $g\in G$, $gN$ generates $G/N$? What is the meaning of $gN\cdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.
 A: Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN=\{gn \, | \, n \in N\}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN \cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.
Now to show that $G/N$ is cyclic you need to find an element of the form $gN \in G/N$ such that it generates every element of $G/N$. In other words, for all $aN \in G/N$, we need to show that there exists a $k \in \Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.  
In the given case, observe that $ m \in \{1,2,3,4\}$ for the matrices to be invertible and
$$\begin{bmatrix}m&b\\0&1\end{bmatrix}=
\begin{bmatrix}m&0\\0&1\end{bmatrix}\color{blue}{\underbrace{\begin{bmatrix}1&bm^{-1}\\0&1\end{bmatrix}}_{\in N}} \in \begin{bmatrix}m&0\\0&1\end{bmatrix}\color{blue}{N}$$
Thus
$$G/N=\left\{\begin{bmatrix}m&0\\0&1\end{bmatrix}N \, \Big|\, m \in \{1,2,3,4\}\right\}.$$
Thus $G/N$ has four elements and now we can see that either $\begin{bmatrix}2&0\\0&1\end{bmatrix}N$ or $\begin{bmatrix}3&0\\0&1\end{bmatrix}N$  can act as generators.
A: Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.
By definition, we have $gN\cdot hN=\{gn\cdot hm:n,m\in N\}$
And, we can see that $gN\cdot hN=\{gh\,h^{-1}nhm:n,m\in N\}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $G\to G/N$ a homomorphism.
And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4\in N$ ($\Leftrightarrow\  g^4N=eN$)
 but $g^2\notin N$.
A: We know it has order $4$, because $\mid G\mid=5^2-5=20$.
To be cyclic means there's an element of order $4$.
How about $\begin{pmatrix}2&1\\0&1\end{pmatrix}N$?
