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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 $cin mathbb{Z}_5$.
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 $gin G$, $gN$ generates $G/N$? What is the meaning of $gNcdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.

Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.

By definition, we have $gNcdot hN={gncdot hm:n,min N}$

And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ 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 $g^4in N$ ($Leftrightarrow g^4N=eN$)
but $g^2notin N$.

We know it has order $4$, because $mid Gmid=5^2-5=20$.

To be cyclic means there's an element of order $4$.

How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?

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
$$begin{bmatrix}m&b\0&1end{bmatrix}=
begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$
Thus
$$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.