"Dumbed Down" Definition Behind Factor Groups and Computing Them I'm confusing myself trying to understand what factor groups really are. I see that the formal definition is:
If $N$ is a normal subgroup of $G$, then the group of left cosets of $N$ form a group under the coset multiplication given by $aN$ $bN$ = $ab N$. This is called the factor group of $G$ determined by $N$, denoted by $G/N.$
If I'm understanding it correctly, then is a factor group simply "just the set of all left cosets of $N$" (as long as $N$ is normal)? If so, then my understanding of computing said $G/N$ would be to:
1) Check $N$ really is a normal subgroup of $G$. (left cosets of $N$ = right cosets of $N$, $N$ satisfies all axioms of a subgroup).
2) Compute all left cosets of $N$. (this is where I'm having some trouble as well; how do you know which cosets are the "correct" left cosets?)
3) Put them all in a set s.t. $S$ = $G/N$. 
Is this correct? I feel like this is way too simple, and couldn't really apply it. For example, I was on a problem that asked:
Compute the factor group ($\mathbb{Z}_4 \times \mathbb{Z}_8$) / <($1, 2$)>.
In this case, how I couldn't really start on finding the left cosets of <($1, 2$)>, or how to put them together as a factor group.
I know this question kind of seems overloaded, but I feel like I'm $so$ close to really hitting that "ding" on factor groups. Any help would be great. Thank you!
 A: Here is perhaps an alternative way to think about it.
The factor group is the group we get when we ignore elements of the normal subgroup $N$. So we want the elements $g$ and $gh$ to be the same if $h\in N$, and we want to be able to multiply elements while preserving this property. You can always define the equivalence relation required for this (the equivalence classes are the left cosets) but in order to multiply them you need $N$ to be normal. Here's why; people make it out to be this really deep concept, but it's silly algebraic manipulation.
Let $a,b\in G$ and $g,h\in N$. We want to be able to say that $ab$ is equivalent to $(ag)(bh)$, in that they are in the same left coset. Rewrite this as
$$(ag)(bh)=ab(b^{-1}gb)h$$
In order for this to be equivalent to $ab$ it must be the case that $(b^{-1}gb)h\in N$. Since $h\in N$, this means that the condition is that $b^{-1}gb\in N$ for all $b\in G$ and $g\in N$. This is very often taken as the definition of a normal subgroup, and this is the reason. 
For the definition you seem to be using, we require that the left cosets be the same as the right cosets so we can move the $g$ past the $b$. We need to know that
$$gb=bg'$$
for some $g'\in N$. This means that
$$(ag)(bh)=abg'h$$
which gets us to the same place, but it's less specific. It happens that $g'=b^{-1}gb$.
Hope this helps.
