the rank 3 free group is embeddable in the rank two free group


Let F=\langle x,y|\ \rangle be the rank two free group and U=\langle\{x^2,y^2,xy\}\rangle be a subgroup.
Observe that xy^{-1}=xy(y^2)^{-1}, then xy^{-1}\in U.

Clearly F=U\sqcup Ux, because it is not difficult to convince oneself that U consists on words of even length and xy^{-1}\in U implies Uy=Ux.

Technically, that is attending to the Schreier’s recipe, having \Sigma=\{1,x\} as a set of transversals and being S=\{x,y\} the free generators for F.

Set \Sigma S=\{x,\ y,\ x^2,\ xy\} and take \overline{\Sigma S}=\{1,x\}, then we get

\overline{\Sigma S}^{-1}=\{1,x^{-1}\}.

So according to Schreier’s language the set  \Sigma S\overline{\Sigma S}^{-1}=\{ gs\overline{gs}^{-1}|g\in\Sigma,s\in S\}, in our case, is

\{\ x\overline{x}^{-1}=1\ ,\ y\overline{y}^{-1}=yx^{-1}\ , \ x^2\overline{x^2}^{-1}=x^2\ ,\ xy\overline{xy}^{-1}=xy\ \}.

Hence \{\ xy^{-1}\ ,\ x^2\ ,\ xy\ \} are the free generator for U.

Note that this three word are the first three length-two-words in the alphabetical order, start by  1<x<x^{-1}<y<y^{-1} and continuing  to

x^2<xy<xy^{-1}<x^{-2}<x^{-1}y<x^{-1}y^{-1}

. . .< yx<yx^{-1}<y^2<y^{-1}x<y^{-1}x^{-1}<y^{-2}

 

9 Comments

Filed under algebra, cucei math, free group, group theory, math, mathematics, what is math, what is mathematics

9 responses to “the rank 3 free group is embeddable in the rank two free group

  1. rrogers31: I recently wrote the little “Schreier’s language” to my Algebra- Moderna-Three ‘s course. Consider keep on asking for more, ‘cuz is enjoyable to see other’s viewpoint.

  2. is it also possible F_4\hookrightarrow F_2 or F_4\hookrightarrow F_3???

    • Am I underestimating the problems involved?
      But:
      Drawing from the Wikipedia page on free groups
      http://en.wikipedia.org/wiki/Free_group
      ; wouldn’t this all follow from “Facts and Theorems ”
      1-3: Two free groups F(S) and F(T) are isomorphic if and only if S and T have the same cardinality. This cardinality is called the rank of the free group F. Thus for every cardinal number k, there is, up to isomorphism, exactly one free group of rank k.
      2-2: A free group of rank k clearly has subgroups of every rank less than k. Less obviously, a (nonabelian!) free group of rank at least 2 has subgroups of all countable ranks.

      • the point rrogers31 2.2) gives you the case F_2\subset F_3. There it says “clearly”. Which I agree.

        But F_3\subset F_2 is counterintuitive but true. That is what I discuss.

        For more evidence of this check link

      • for a review of Schreier ideas follow to this.

      • “Less obviously, a free group of rank at least 2 has subgroups of all countable ranks” covers the F_k\subset F_2, rrogers31 :D

      • “for a review of Schreier ideas follow to this.”
        Thanks for the trouble of writing this up! I will study it. In my case it’s not a review; while my Abstract Algebra class (s) covered free groups they didn’t cover this.

      • Back again (:
        When you have time, can you take a look at the Free-discussion link on.
        http://pascal-umbral.blogspot.com/
        Either the last post or under “My Papers”.
        And comment someway :)
        Sorry to clutter this page (which I was trying to avoid); just errase this and the last two comments.

  3. Let me see if I have this right; I’m a little slow.
    Your are saying: let the final U be the redesignated elements of the free group a,b,c.
    Then certainly any element u of U can be mapped by u-a to an element f of F by using your last statement.
    The question is: Do every distinct two elements of U (written as abc.. then xy…) correspond to distinct elements of F?
    You answer in the affirmative!
    and in addition that it is homomorphic in the sense of http://en.wikipedia.org/wiki/Embedding#Algebra: Universal algebra and model theory;
    characterization line 2. (sorry but I didn’t feel like reproducing the entire section).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s