$\Bbb Z$ is the group of integers. It is abelian and cyclic. Abelian because for any two integers $a, b$, we have $a+b = b+a$. Cyclic because you can pick an element $c$ (either $-1$ or $1$, it doesn't matter) such that every element other than $0$ can either be written $c+c+\cdots c$ or as the inverse of such a number. Any such choice of $c$ is called a generator for the cyclic group $\Bbb Z$. When a group is not cyclic, you need more than one generator at the same time to reach every element.
The group $\Bbb Z\oplus \Bbb Z$ has things like $(5, 4)$ as elements. The group operation works in a way that is called "component-wise". We have, for instance, $(5, 4) + (2, 7) = (5+2, 4+7) = (7, 11)$. As you can see, the first components $5$ and $2$ are added, and so are the second components $4$ and $7$. There is no "interaction" between the first and second component at this level. This is still abelian, but it is no longer cyclic. You may for instance pick $(1, 0)$ and $(0, 1)$ as generators, but there are many other choices.
Now we get to quotient groups. In a quotient group, we look at the "numerator group" (base group) (in this case $\Bbb Z\oplus \Bbb Z$), and we "forget" some of the structure that is there. This means that there are some things that are not equal in the base group (the "structure" of the base group can tell those elements apart), but they are equal in the quotient group (they cannot be told apart). The criterion for not being able to tell them apart is that you can get from one to the other by only applying elements from the "denominator" group. This will work nicely if and only if the denominator group is a normal subgroup (in an abelian group, all subgroups are normal, so this is not an issue here).
One very simple example is $\Bbb Z\oplus \Bbb Z / \langle(0, 1)\rangle$. An example element in this group can be written as $(-3, 5) + \langle(0, 1)\rangle$ (this is exactly the $aH$-notation, using addition instead of multiplication). One other way of writing the exact same element in the is $(-3, -1) + \langle(0, 1)\rangle$. These are considered equal because in the base group you can get from $(-3, 5)$ to $(-3, -1)$ by adding $(0, -6)$, which is inside the denominator group $\langle(0, 1)\rangle$. We say that $(-3, 5)$ and $(-3, -1)$ represent the same element in the quotient group.
The net effect of taking this quotient is that we "forget" that the last component even exists. This quotient group is cyclic, with generator $(1, 0) + \langle(0, 1)\rangle$.
Lastly, we get to your group $\Bbb Z\oplus \Bbb Z/\langle(4, 2)\rangle$. For each element in this group, there is exactly one representative with second coordinate equal to either $0$ or $1$. For instance, $(-8, 6)$ represents the same element as $(-32, 0)$. In other words, $(-8, 6) \sim (-32, 0)$. In other words
$$
(-8, 6) + \langle(4, 2)\rangle = (-32, 0) + \langle(4, 2)\rangle
$$
Note that the elements are equal in the quotient group, but only equivalent in the base group. This is an important distinction.
That being out of the way, there are infinitely many elements in $\Bbb Z\oplus \Bbb Z$ with second coordinate equal to $0$ or $1$, and they cannot be equivalent. Therefore there are infinitely many elements in the quotient group. The quotient group is also not cyclic. The argument is below.
Note that in the quotient group we still have some sense of whether the first and second components are odd or even. There are now three kinds of elements that cannot all three be reached by any single generator:
- An element with odd first component and even second component
- An element with even first component and odd second component
- An element with both components odd
(Actually, any two of these would be enough.) Therefore the group cannot be cyclic.
Best Answer
Let's look at a slightly more "in-depth" example: Consider the subgroup $7\Bbb Z$ of $\Bbb Z$ which consists of all (positive, negative and $0$) multiples of $7$. Let's "expand" this a bit:
$7\Bbb Z = \{\dots,-21,-14,-7,0,7,14,21,28,\dots\}$
What set do we get if we add $1$ to everything in this set?
$1 + 7\Bbb Z = \{\dots,-20,-13,-6,1,8,15,22,29,\dots\}$
If we shift "over $2$" we get:
$2 + 7\Bbb Z = \{\dots,-20,-12,-5,2,9,16,23,30,\dots\}$
Now here is a curious thing: If we take something in, say $3 + 7\Bbb Z$, and add it to something in $2 + 7\Bbb Z$, we'll get something in $5 + 7\Bbb Z $. For example:
$10 + 9 = 19$ and $19$ is $5$ more than $14$.
This process "chops up" $\Bbb Z$ into $7$ pieces (each of which is "infinite") and in every piece, the numbers are all "$7$ apart".
So the "$k$" in $k + 7\Bbb Z$ measures how far to the right (right = counting up) we are from a multiple of $7$.
If you imagine the integers as an infinite string of beads, we are "wrapping the beads into a circle" so that the $7$th bead winds up in the same place as the "$0$-bead". In fact, any two beads that are a multiple of $7$ apart on our original string, wind up in the same place on our "$7$ position circle".
You've seen this before: on a clock (there it's "modulo $12$" instead of $7$, or "modulo $24$" if you use military time).
This has the effect of effectively "setting all multiples of $7$ equal to $0$". Since we know that $14$ (for example) isn't actually $0$, we don't say:
$14 = 0$, but rather, $14$ is equivalent to $0$ since it is a multiple of $7$ away from $0$ (if $6$ was "as high as we could count before starting over", $14$ would actually BE $\ 0$).
It's somewhat "magical" that all this actually WORKS, in that we don't get any contradictions or confusion this way. Part of this has to do with the fact that addition is commutative: that is, that $k + m = m+k$ for any two integers $k,m$. With a group in general, we need a more special condition on the subgroup to make this all work out (it has to be normal).
It turns out that with a general group $G$ and a subgroup $H$, that the set:
$(aH)(bH) = \{(ah)(bh'): h,h' \in H\}$ doesn't always equal some other coset $kH$, so subgroups for which this DOES happen are "special".
If $G$ is abelian, of course, then $(ah)(bh') = a(hb)h' = a(bh)h' = (ab)(hh')$ and this DOES happen. So quotient groups of abelian groups are "easier to understand".
In this special case of $G = \Bbb Z$ and $H = ${even integers}, this process is called "the arithmetic of parity" (reasoning by evens and odds). This "trick" (reducing mod $2$) turns an "infinite number of cases" to just TWO cases, and often that is all we need.