# Tag Archives: free groups

## 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