# Category Archives: free group

modern medular math concept

## transversal rewriting solution by semidirect product of certain coset maps

este proceso se generaliza

2014/08/21 · 13:48

## producto semi-directo

diagram chasing the wreath

2014/05/03 · 14:50

## permutational wreath product

Having an action $G\times R\to R$ between two groups means a map $(g,r)\mapsto ^g\!r$ that comply

• ${^1}r=r$
• $^{xy}r=\ ^x(^yr)$
• $^x(rs)=\ ^xr ^xs$

Then one can assemble a new operation on $R\times G$ to construct the semidirect product $R\rtimes G$. The group obtained is by operating

$(r,g)(s,h)=(r\ {^h}s,g\ h).$

Let $\Sigma$ be a set and $A^{\Sigma}$ the set of all maps $\Sigma\to A$. If we have an action $\Sigma\times G\to\Sigma$ then, we also can give action $G\times A^{\Sigma}\to A^{\Sigma}$ via

$gf(x)=f(xg)$

Then we define

$A\wr_{\Sigma}G=A^{\Sigma}\rtimes G$

the so called permutational wreath product.

This ultra-algebraic construction allow to give a proof  of two pillars theorems in group theory: Nielsen – Schreier and Kurosh.

The proof becomes functorial due the properties of this wreath product.

The following diagram is to be exploited

Ribes – Steinberg 2008

Filed under algebra, free group, group theory, math, maths, what is math

## 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

## words-length and words in a group

the group is $\mathbb{Z}_2*\mathbb{Z}_3=\langle a,\ b\mid a^2,\ b^3\rangle$, and the picture:

Correctly predicts the next one. It is A164001 in the OEIS data-base. Dubbed “Spiral of triangles around a hexagon“. It has the generating function $-(x+1)+\frac{x^2+2x+1}{1-x^2-x^3}$, Why would it be? :) . This another is A000931.

Filed under free group, math, sum of reciprocals

## is it enough…

that the presentation given by

$\langle A,B\ :\ A^2=B^3,\quad A^2B=BA^2,\quad A^4=e,\quad B^6=e\rangle$

determines the group $SL_2({\mathbb{Z}}/{3\mathbb{Z}})$?

Similar question for

$\langle A,B,C$:

$A^2=B^3$,

$A^2B=BA^2$,

$A^4=B^6= C^2=e$,

$AC=CA,BC=CB\rangle$

for the group $GL_2({\mathbb{Z}}/{3\mathbb{Z}})$.

Other similar problems but less “difficult” are:

• $\langle\varnothing:\varnothing\rangle=\{e\}$
• $\langle A\ :\ A^2=e\rangle$  for  $\mathbb{Z}_2$
• $\langle A\ :\ A^3=e\rangle$  for  $\mathbb{Z}_3$
• $\langle A\ :\ A^4=e\rangle$  for  $\mathbb{Z}_4$
• $\langle A,B\ :\ A^2=e, B^2=e, AB=BA\rangle$  for  $\mathbb{Z}_2\oplus\mathbb{Z}_2$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=B^2A\rangle$  for  $S_3$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=BA\rangle$  for  $\mathbb{Z}_2\oplus\mathbb{Z}_3$
• $\langle A:\varnothing\rangle$ it is $\mathbb{Z}$
• $\langle A,B:AB=BA\rangle$  is $\mathbb{Z}\oplus\mathbb{Z}$
• $\langle A,B:\varnothing\rangle$  is $\mathbb{Z}*\mathbb{Z}$, the rank two free group
• $\langle A,B,J:(ABA)^4=e,ABA=BAB\rangle$ for $SL_2(\mathbb{Z})$
• $\langle A,B,J:(ABA)^4=J^2=e,ABA=BAB, JAJA=JBJB=e \rangle$ for $GL_2(\mathbb{Z})$
• $\langle A,B:A^2=B^3=e \rangle$ for $P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3$
• dare altri venti esempi
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