TITLE: Equivalence relation on a group and normal subgroups
QUESTION [5 upvotes]: Let $G$ be a group and define an equivalence relation $R$ on it. Let $G/R$ be the set of equivalence classes of this relation. Then is $G/R$ equal to $G/N$ for some normal subgroup $N$ of $G$?
Sorry, not even sure if the question makes sense. I am trying to basically ask if there is a connection between any equivalence relation defined on a group and normal subgroups of $G$? Or rather, does an equivalence relationship give rise to a normal subgroup? Thank you.

REPLY [3 votes]: An interesting connection is the notion of the $presentation $ of a group.  Given generators $S$ and relations $R$, one can consider the free group on the generators modulo the $\bf{normal\, subgroup\,generated\,by}$ the relations.  This is denoted $\langle S|R\rangle$.
It can be shown that every group has such a presentation.  But in general the presentation is not unique.
An example is the cyclic group $C_n=\langle x|x^n\rangle$.

As to your question, $G/R$ may not be a group.  Given a subgroup $H$, there is a natural way to define an equivalence relation on $G$.  Namely, $a\sim b \iff ab^{-1}\in H$.  $G/H$ won't be a group, unless $H$ is normal.  There is the notion of the normal closure of $H$, the smallest normal subgroup containing $H$, alluded to above.