Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factor (Quotient) Space definitions.

  1. Dec 28, 2013 #1
    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 [itex]G[/itex] and a group [itex]G'[/itex] and there exists a homomorphism [itex]R: G → G'[/itex] and for any element [itex]g \in G[/itex], the equivalence class of g is denoted as [itex][g]_{R} = \{h \in G \:|\: f(h) = f(g)\}[/itex]

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

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

    [itex]G/H = \{gH \:|\: g \in G\}[/itex]

    How are these definitions stating the same thing? Does it have something to do with [itex]H[/itex] being the kernel of a homomorphism? I don't really understand what cosets have to do with equivalence relations.
     
  2. jcsd
  3. Dec 28, 2013 #2
    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
    [tex]f(g) = f(h)~\Leftrightarrow f(g^{-1}h) = e~\Leftrightarrow g^{-1}h\in H~\Leftrightarrow g\sim h[/tex]
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Factor (Quotient) Space definitions.
  1. Quotient Spaces (Replies: 8)

  2. Quotient Spaces (Replies: 2)

Loading...