How to find all elements of $\mathbb{Z}_{4} \times\mathbb{Z}_{4}/\langle(1,1)\rangle$? I am studying factor groups, and I saw an example that says

Find all the elements of the factor group $\mathbb{Z}_{4}   \times \mathbb{Z}_{4}/\langle(1,1)\rangle$. 

I know that the order of  $\mathbb{Z}_{4}  \times \mathbb{Z}_{4}$  is $16$, and the order of $\langle(1,1)\rangle$ is $4$, so we must have $4$ elements, but how can I find them? 
Edit: Actually, this is not a homework, I have the answer saying that those elements are  $\{(1,1),(2,2),(3,3),(0,0)\} , \{(1,2),(2,3),(3,0),(0,1)\} , \{(2,1),(3,2),(0,3),(1,0)\},\{(1,3),(2,0),(3,1),(0,2)\}$. 
But I did not understand how they are found, can anyone please explain the steps to find them?
Thanks.
 A: In general, if $H$ is a normal subgroup of $G$, the elements of the quotient group $G/H$ are by definition the cosets $gH$ for $g\in G$. Here, $H = \langle(1,1)\rangle$.
To find the cosets $gH$, write them one at a time. First, we know we have $$H = \{(1,1),(2,2),(3,3),(0,0)\}$$ Now take an element in $\mathbb Z_4 \times \mathbb Z_4$ but not in $H$. Let's say $(0,1)$. This gives us the coset $$(0, 1) + H = \{(1,2),(2,3),(3,0),(0,1)\}$$ Again, we pick a new element, this time one that is in none of the cosets we currently have. Let's take $(1,0)$ which gives us the coset $$(1,0) + H = \{(2,1),(3,2),(0,3),(1,0)\}$$ Finally, we need to pick one last element not in any of these cosets. Let's pick $(0,2)$ giving us the coset $$(0,2) + H = \{(0,3),(2,0),(3,1),(0,2)\}$$
Now we have the $4$ elements of the quotient.
A: HINT 1 The elements of the factor group are only uniquely presented up to an element of $\langle 1,1\rangle$, meaning that $(a,b)$ and $(a,b)+(1,1)$ represent the same element in the factor group.
HINT 2 When are two elements $(a,b)$ and $(c,d)$ different in the factor group? (precisely when $(a-c,b-d) \neq c(1,1)$ for some $c$)
A: Note that taking a quotient by $\langle 1,1\rangle$ takes any element of the form $(x,x)$ to the identity coset.  Why not look at an element like $(1,0)$?
A: Just take a second linearly independent and indivisible element of your group and look what it generates. The elements in the generated group will be a solution to your problem. 
