How to find all elements of a factor group? I am trying to understand the meaning of a factor group. I know how to find the order of a factor group, for example, if I try to find the order of $\mathbb{Z}_4\times\mathbb{Z}_4/\langle(1,1)\rangle$, it is $\frac{16}{4} = 4$. But, how do I find all the elements of a factor group, and what is the meaning of a factor group? I know the definition but what does it mean intuitively?
Actually, I have found the elements like the following: 
\begin{align}
\{(1,1),(2,2),(3,3),(0,0)\} , \{(1,2),(2,3),(3,0),(0,1)\}, \\
\{(2,1),(3,2),(0,3),(1,0)\} , \{(1,3),(2,0),(3,1),(0,2)\}
\end{align}
and I guess this is true.
My other question is that, the question says that this group is isomorphic to some other group, how can I decide which group is isomorphic to it? Since it has 4 elements, can I say that it is isomorphic to Z4? 
Thank you
 A: Yes, what you have done is correct.  And you can in fact say that your group is isomorphic to $Z_4$.  Here's one way. 
Let $G = Z_4 \times Z_4 /\langle (1,1) \rangle$.  Define a map $\phi : G \to Z_4$ by the rule 
$$
\phi([[x] \times [y]]) = [y-x].
$$
Now show that $G$ is an isomorphism of groups. 
As to understanding the group intuitively, I can try to explain my viewpoint.  In the lattice $\mathbb{Z} \times \mathbb{Z}$ draw the sublattice generated by the elements $(4,0)$ and $(0,4)$.  This should look like a collection of dots, all evenly spaced.  Imagine shifting the collection of dots up by one unit.  This collection of shifted dots represents the element $[0] \times [1]$ in $\mathbb{Z}/4 \times \mathbb{Z}/4$.  Similarly, by shifting our original collection of dots to the right by one unit gives the element $[1] \times [0]$ in $\mathbb{Z}/4 \times \mathbb{Z}/4$.   
Now that we understand $\mathbb{Z}/4 \times \mathbb{Z}/4$, let us try to understand the group $G$ (defined above).  Consider the four sublattices $[0] \times [0]$, $[1] \times [1]$, $[2] \times [2]$, and $[3] \times [3]$.  These all belong to the ideal $\langle (1,1) \rangle$ generated by $[1] \times [1]$. This collection of sublattices is now identified as a single group element, call it $\overline{0}$.  Now imagine shifting $\overline{0}$ to the right by one unit.  This newly shifted collection of sublattices is the single group element we call $\overline{1}$. We shift again to the right to obtain $\overline{2}$.  We see that we will obtain four distinct collections of sublattices in this way.  They are $\overline{0}, \overline{1}, \overline{2}, \overline{3}$.  Notice that the group we obtained was generated by a single element, so our group is cyclic and of order 4. 
