Equivalence relation on a group and normal subgroups

Question

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.

Answer 1

An arbitrary equivalence relation $\sim$ not satisfying any extra conditions won't behave like this; for example, you can pick an equivalence relation such that the equivalence classes don't have the same size.

The correct thing to do is to impose an axiom saying that the group structure on $G$ descends to the quotient $G/\sim$. If you spell this out what it means is that if $g_1 \sim g_2$ and $h_1 \sim h_2$ then $g_1 h_1 \sim g_2 h_2$, or equivalently that the subset

$$\{ (g_1, g_2) \in G \times G : g_1 \sim g_2 \}$$

is a subgroup of $G \times G$. And then you can show that there's a natural bijection between equivalence relations satisfying this property (congruences in the category of groups) and normal subgroups of $G$, where the correspondence goes:

1. Given a congruence $\sim$, the normal subgroup is the subgroup $\{ g \in G : e \sim g \}$.
2. Given a normal subgroup $N$, the congruence is $g \sim h \Leftrightarrow \exists n \in N: g = hn$.

Furthermore this bijection respects quotients in the sense that if $\sim$ corresponds to $N$ then $(G/\sim) \cong G/N$. See this blog post for details. This is, in my opinion, the correct way to introduce normal subgroups to students.

(Going further: exactly the same idea works for rings and gives a bijection between congruences on rings and two-sided ideals. For monoids we really do just have to work with congruences and there's no longer a notion of "normal submonoid" or "ideal of a monoid.")

Answer 2

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 $C_n=\langle x|x^n\rangle$.

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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.