TITLE: Quotienting a group by an equivalence relation
QUESTION [6 upvotes]: In Jacobson's book, Basic Algebra 1, he points out that you can form quotient groups by looking at the equivalence classes of a 'multiplication compatible' equivalence relation on the group. By multiplication compatible, he means that if $A,B$ are equivalence classes for $a, b$, and we also have that $c \in A, d \in B$, then the equivalence class of $ab$ is the equivalence class of $cd$. As it turns out, this forms a group. He then points out that the equivalence class of 1 is a normal subgroup and that the group that you obtained is equal to the original group quotiented by the equivalence class of 1. I'm curious as to if there are any situations where you would quotient a group by an equivalence relation, and not by a normal subgroup, given that these are equivalent.

REPLY [5 votes]: Sure. Consider the quotient of a topological group $G$ by the relation "lies in the same connected component." You get a group $\pi_0(G)$ of connected components. This is equivalent to quotienting by the connected component of the identity (which incidentally proves that this subgroup is normal) but I think the equivalence relation is the more natural thing to think about, even if thinking about the normal subgroup is convenient.