# Quotienting a group by an equivalence relation

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.

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The image of a homomorphism $f$ is the quotient by the obvious equivalence: $x\equiv y$ iff $f(x) = f(y)$. That the image is isomorphic to this quotient is tautological, but is also called the first isomorphism theorem. –  Jack Schmidt Aug 3 '12 at 16:57
Modular arithmetic definitions often use $a \equiv b \mod n$ iff $n \mathop{|} (a-b)$, which is basically the normal subgroup version. However if $n=10$, then to check if $398743$ is equivalent to $984938$ you don't need to subtract, you just use the last digit: $3 \neq 8$ so they are not equivalent. This is the use of a direct equivalence relation, $x\equiv y$ iff the last digits of $x$ and $y$ agree. –  Jack Schmidt Aug 3 '12 at 17:04
@Jack That's a good way of thinking about it, thanks. –  Chris Dugale Aug 3 '12 at 17:17
@jack (2nd post). Yeah, I got that motivation, but I was just wondering if you would use this method in any other settings. –  Chris Dugale Aug 3 '12 at 17:18
the problem is $g h^{-1}$ is not much harder to compute than just looking at $g$ and $h$. However for a matrix group, it is conceivable that you could have an $O(n)$ algorithm to compute $g \equiv h$, while $g h^{-1}$ is $O(n^3)$. Similarly, multiplication in a polyclically presented group is sub-exponential runtime, but check membership in one of the subnormal subgroups used in the definition of the presentation is $O(n)$. This is actually used in calculations with finite soluble groups. –  Jack Schmidt Aug 3 '12 at 17:24

Another type of equivalence relation you see in group theory has to do with pairs of subgroups, rather than elements. If $1\leq M\trianglelefteq H \leq G$, then $(H,M)$ is referred to as a pair if $H/M$ is cyclic. $(H,M)$ is called "good" if $[g,H\cap g H g^{-1}]\not\subseteq M$ for any $g\in G\setminus H$. Lastly, if $(K,L)$ is another good pair, we call $(H,M)$ and $(K,L)$ related in G if there is a $k\in G$ so that $k^{-1}Hk\cap L=K\cap k^{-1}Mk$. (There are character theoretic definitions for all of these, too.) As it turns out, being "related" is an equivalence relation on the set of good pairs in G.
I don't know whether the second example is what you were looking for exactly when you talk about taking a quotient, but the interactions between these equivalency classes can be very interesting, and ends up motivating the study of a complicated class of groups called $M$-groups.
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.