# 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 and continuing  to $x^2 $. . .< yx

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

1. juanmarqz

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. juanmarqz

is it also possible $F_4\hookrightarrow F_2$ or $F_4\hookrightarrow F_3$???

• rrogers31

Am I underestimating the problems involved?
But:
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.

• juanmarqz

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

• juanmarqz

for a review of Schreier ideas follow to this.

• juanmarqz

“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

• rrogers31

“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.

• rrogers31

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. rrogers31

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?