Factor (Quotient) Space definitions.

[wotanub]

I'm learning algebra by myself and this concept is confusing me. Please excuse me if I define anything wrong... I've never expressed myself in this language before.

Lets say we have a group $G$ and a group $G'$ and there exists a homomorphism $R: G → G'$ and for any element $g \in G$, the equivalence class of g is denoted as $[g]_{R} = \{h \in G \:|\: f(h) = f(g)\}$

I understand the factor space $G/R$ as the set of all equivalence classes of $G$:
$G/R = \{[g]_{R} \:|\: g \in G\}$

but another way I always see this explained (that I'm not clear on) is if we have a subgroup $H \subset G$ then we can define a factor space with left cosets.

$G/H = \{gH \:|\: g \in G\}$

How are these definitions stating the same thing? Does it have something to do with $H$ being the kernel of a homomorphism? I don't really understand what cosets have to do with equivalence relations.

 

[R136a1]

The two are closely linked.

Take the factor group approach. The quotient set is defined as $G/H = \{gH~\vert~g\in H\}$ with $H$ a subgroup. This $H$ has to be normal if you want $G/H$ to have a natural group structure, so we will do this. This actually corresponds to the following equivalence relation: we define $g\sim h$ iff $g^{-1}h\in H$. The equivalence classes correspond exactly to the cosets. That is: $[g]_\sim = gH$. So the coset definition actually does correspond to an equivalence relation.

Now, the link with homomorphisms is the following:
Given a homomorphism $f:G\rightarrow G^\prime$, then we set $gRh$ iff $f(g) = f(h)$. But we can take $H = \textrm{Ker}(f) = \{h\in G~\vert~ f(h) = e\}$. This is a normal subgroup. Then we see that
$$f(g) = f(h)~\Leftrightarrow f(g^{-1}h) = e~\Leftrightarrow g^{-1}h\in H~\Leftrightarrow g\sim h$$
So this equivalence relation is nothing more than the one defined above.

Conversely, if we are given a normal subgroup $H$ of $G$, then we can always find a group $G^\prime$ and a homomorphism $f:G\rightarrow G^\prime$ such that $H = \textrm{Ker}(f)$. Indeed, just take $G^\prime = G/H$ and take $f(g) = gH$.

So the two methods outlined by you are equivalent.