Computing the order and cyclicity of quotients of direct products. 
Determine the order of $(\mathbb{Z} \times \mathbb{Z} )/ \langle(2,2)\rangle$ and $(\mathbb{Z} \times \mathbb{Z} )/ \langle(4,2)\rangle$. Are the groups cyclic?

I've read many solutions on the internet including here (see 20\187) and here as well as the answer in my book. None of them made sense to me since they all seem to have contradictory methods and answers.
In general, what are people doing when they are trying to find the order by adding $(0,1), (1,0)$ or some variation of those elements to the generators? Could someone work out this one problem for me (with explanation of each thought process)? I am getting tired of seeing only shortcuts and solutions with no justification for their steps.
 A: Let's start with a simpler case: $\mathbb{Z}/2\mathbb{Z}$.  Here, $2\mathbb{Z}=\{\ldots,-4,-2,0,2,4,\ldots\}$.  So, the cosets of $2\mathbb{Z}$ in $\mathbb{Z}$ are the $z+2\mathbb{Z}=\{\ldots,z-4,z-2,z,z+2,z+4,\ldots\}$.  If that $z$ would happen to be $2$, we'd get back $2\mathbb{Z}$ (just shifted over one).  Thus $2+2\mathbb{Z}=2\mathbb{Z}$.  Since you can write any number as $2m+n$ where $m$ is some integer and $n$ is $0$ or $1$, we can characterize all cosets of $\mathbb{Z}/2\mathbb{Z}$ by $2m+n+2\mathbb{Z}=n+2m+2\mathbb{Z}=n+2\mathbb{Z}$, so the only two cosets are $2\mathbb{Z}$ and $1+2\mathbb{Z}$.
Now, keeping that process in mind, let's do $(\mathbb{Z}\times\mathbb{Z})/\langle(2,2)\rangle$.  We have  $$\langle (2, 2 ) \rangle =\{\ldots,(-4,-4),(-2,-2),(0,0),(2,2),(4,4),\ldots\}$$
Now if somebody adds $(1,0)$ to each of these numbers, we get $$(1,0)+\langle (2, 2 ) \rangle =\{\ldots,(-3,-4),(-1,-2),(1,0),(3,2),(5,4),\ldots\}$$ If we keep adding them, we will keep shifting the left and right numbers apart: they'll never equal $\langle (2, 2 ) \rangle$ again because nothing modifies the second coordinate.  Thus, the group is infinite.
Now, is it cyclic?  It's infinite, any any infinite cyclic group is isomorphic to $\mathbb{Z}$.  This is where the part where you can add $(1,1)+\langle(2, 2 )\rangle$.  You can easily verify this has order $2$ in just the same way as the $2\mathbb{Z}$ example, and $\mathbb{Z}$ contains no elements of finite order, so it must not be cyclic.
A: I think you are misreading the first link. They are not adding elements to generators. When $G$ is a group and $H$ a normal subgroup then $G/H$ is a set of cosets. In the case of $G = \mathbb Z$ and $H = \langle (2,2) \rangle$ the elements in $G/H$ are written $(x,y) +  \langle (2,2) \rangle$ and in general this is written $g + H \in G/H$.
To consider an easier example: Let $G = \mathbb Z$ and $H = 2 \mathbb Z$. Then $G/H = \mathbb Z / 2 \mathbb Z$. What do the elements look like? We identify two elements in $G$ if they differ by an element in $H$. Let's start by considering $0$: elements that get identified with zero are $0,2,4,6,8 \dots$, that is, all even numbers stand for zero in $\mathbb Z / 2 \mathbb Z$. Let's denote this element by $\bar{0}$. Next let's see what happens to $1$: elements that get identified with $1$ are $1,3,5,7,9,11,13,15,\dots$, that is, all odd integers. Let's denote this element $\bar{1}$. Since odds and evens make up all of $\mathbb Z$ we are already done: $\mathbb Z / 2 \mathbb Z$ looks like $\{\bar{0}, \bar{1}\}$. Alternatively, one can denote them $\bar{0} = 0 + 2\mathbb Z$ and $1 + 2 \mathbb Z$.
Now let's try to find out what $(\mathbb{Z} \times \mathbb{Z} )/ \langle(2,2)\rangle$ looks like. Two elements in $\mathbb Z \times \mathbb Z $ are the same if they differ by a multiple of $(2,2)$. The elements that get identified with $(0,0)$ are $(0,0), (2,2), (4,4), (6,6), \dots$ that is, $(2k,2k)$ for all $k$ in $\mathbb Z$. Let's denote that $\overline{(0,0)}$. You can't proceed in this way and find all the elements of $(\mathbb{Z} \times \mathbb{Z} )/ \langle(2,2)\rangle$ in this case because, as you know form the first link, this group is infinite. As pointed out in the link, $\overline{(n,0)}$ and $\overline{(m,0)}$ are distinct elements of the quotient group (they are "distinct cosets") for $n \neq m \in \mathbb Z$.  (Perhaps it would be good if you tried to show that cosets are either distinct or equal, as an exercise. ) Now we know that $(\mathbb{Z} \times \mathbb{Z} )/ \langle(2,2)\rangle$ has infinitely many elements. 
It remains to think about whether it is cyclic or not. Hope this helps.
