A generalization of the quotient group: an exhaustive analysis of all possible cases

We know that if $G$ is a group and $N$ is a normal subgroup of $G$, then $G/N$ is a quotient group.

Today, we shall try and explore some fundamental questions that have plagued my understanding of Algebra for a long long time.

What if $N$ is just a subset of $G$, and not necessarily a subgroup? Does $G/N$ mean anything then? Take $G=\Bbb{Z}$ and $N=\{1,2,3\}$. Then $\{g+N| \forall g\in G\}$ consists of $\overline{a}$ for all $a\in G$. For two different $a,b\in G$, we have $\overline{a}\neq\overline{b}$. Hence, there exists a bijection $f:G\to G/N$.

Can $G/N$ display properties of a group? Sure. Define a new operation on $G/N: \overline{a}+\overline{b}=\overline{a+b}$. We have a two-sided identity, a two-sided inverse, associativity, and algebraic closure. We have a group!!

What happens when $N$ becomes a subgroup? For any $n\in N$, we see that $\overline{a+n}=\overline{a}$. Imagine that for each equivalence class, we fix a representation; i.e. for the equivalence class $\{a+n_1,a+n_2,\dots\}$, we choose the element $a+n_1$ for all arithmetic operations, without the need to call any other element for any further operations. Let $\overline{a+n_1}$ and $\overline{b+n_2}$ be the representative elements of two equivalence groups in $G/N$. What about $\overline{a+n_1}+\overline{b+n_2}$? Can we prove that we’ve chosen $\overline{a+n_1+b+n_2}$ to be the representative element of some equivalence class? Probably not.

Now let $N$ be a normal subgroup. Here $\overline{a+n_1+b+n_2}=\overline{a+b+n_3+n_2}$. So we have $\overline{a+b+n_4}$, where $n_4=n_3+n_2\in N$. However, we can’t be sure of choosing this element as the representative of some equivalence class either.

Instead of choosing representative elements, let us try a different approach. Let us consider whole cosets $a+N$ at once for $a\in G$. Choosing representative elements becomes irrelevant here. For group operations, we’d have to calculate $(a+N)+(b+N)$; i.e. add each element of $a+N$ to each and every element of $b+N$.

If $N$ is normal, then we have $(a+N)+(b+N)=(a+b+N)$, which surely is an element of $G/N$. Checking for other group properties, we soon see that $G/N$ is a group.

What if $N$ is not normal? Algebraic closure may be violated. Check out this link: http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only

If $N$ is a normal subgroup, then $(a+N)+(b+N)=(a+b+N)$. Note that this isn’t a necessary condition for $G/N$ to be a group As long as $(a+N)+(b+N)=(c+N)$ for some $c\in G$, we would still have algebraic closure. This is just an additional luxury that provides for notational convenience.

Conclusion: It is possible for $G/N$ to be a group if $N$ is not a subgroup at all. The group operation would be $(a+N)+(b+N)=(a+b+N)$. Remember that this would be a result of using the operation $*_G$ on the elements of $(a+N)$ and $(b+N)$. The operation on $G/N$ would just have to be defined this way. Choosing representatives doesn’t even come up here as there is a different $a+N$ for every $a\in G$.

When $N$ is a subgroup, there is one coset for multiple elements in $G$. Hence, choosing representatives comes up. But this proves to be problematic whilst trying to satisfy algebraic closure. Hence, we deal with adding whole cosets to each other. Choosing representatives becomes irrelevant here. For the purpose of satisfying algebraic closure, we find that if $N$ is a normal subgroup, $G/N$ is a subgroup in every instance.