Determine the order of (Z $\bigoplus$ Z)/<(4,2)>. Is the group cyclic?
-A subgroup H of a group G is normal if and only if aH=Ha for all a $\in$ G. $Z_{n}$ is an Abelian and cyclic group. If H is a normal subgroup then the operations on the cosets of H, (aH)(bH)=abH is well-defined, and the cosets of H form a group under this operation.
-Note really sure where to go with this.
Best Answer
$\Bbb Z$ 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:
(Actually, any two of these would be enough.) Therefore the group cannot be cyclic.