## 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

## 2013 in review

The WordPress.com stats helper monkeys prepared a 2013 annual report for this blog.

Here’s an excerpt:

A New York City subway train holds 1,200 people. This blog was viewed about 5,200 times in 2013. If it were a NYC subway train, it would take about 4 trips to carry that many people.

Filed under 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

## cf. Frobenius forms

different companion matrices

Filed under algebra, math, mathematics, what is math, what is mathematics

## double coset counting formula

the double coset counting formula is a relation inter double cosets $HaK$, where $a\in G$ and $H,K$ subgroups in $G$. This is:

$\#(HaK)=\frac{|H||K|}{|H\cap aKa^{-1}|}$

The proof is easy.

One is to be bounded to the study of the natural map $H\times K\stackrel{\phi_a}\to HaK$. And it uses the second abstraction lemma.

The formula allows you to see the kinds of subgroups of arbitrary $H$ versus $K$ a $p-SS$ of $G$, $p-SS$ for the set of the $p$- Sylow subgroups.

Or, you can see that through the action $H\times G/K\to G/K$ via $h\cdot aK=haK$ you can get:

• ${\rm Orb}_H(aK)=HaK$, and
• ${\rm St}_H(aK)=H\cap aKa^{-1}$

then you can deduce:

$|G|=\sum_a\frac{|H||K|}{|H\cap aKa^{-1}|}$

Now, let us use those ideas to prove the next statement:

Let $G$ be a finite group, with cardinal $|G|=q_1^{n_1}q_2^{n_2}\cdots q_t^{n_t}$, where each $q_i$ are primes with $q_1 and $n_i$ positive integers.

Let $H$ be a subgroup of $|G|$ of index $[G:H]=q_1$.

Then, $H$ is normal.

Proof:

By employing $K=H$ in the double coset partition, one get the decomposition:

$G=HeH\sqcup Ha_1H\sqcup...\sqcup Ha_tH$

So by the double coset counting formula you arrive to:

$|G/H|=1+[H:H\cap a_1Ha_1^{-1}]+\cdots+[H:H\cap a_tHa_t^{-1}]$

i.e.

$q_1=1+\frac{|H|}{|H\cap a_1Ha_1^{-1}|}+\cdots+\frac{|H|}{|H\cap a_tHa_t^{-1}|}$

From this, we get $\frac{|H|}{|H\cap a_iHa_i^{-1}|}.

But $|G|=q_1|H|$ as well $|H|=|H\cap H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$ so

$|G|=q_1|H\cap H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$, i.e.

$[H:H\cap a_iHa_i^{-1}]$ divides $|G|$

Then $[H:H\cap a_iHa_i^{-1}]=1$. So $|H|=|H\cap a_iHa_i^{-1}|$… This implies $H=H\cap a_iHa_i^{-1}$ and so $H=a_iHa_i^{-1}$ for all the posible $a_i$, i.e, $H$ is normal.

QED.

## las básicas

estas son las matemáticas antes llamadas “puras”

o no? :D

Filed under math, topology, what is mathematics

## puntos críticos de una función suave en el círculo

En esta breve nota demostraremos que cada función $f:S^1\to{\mathbb{R}}^1$ que tenga un punto crítico aislado debe de tener otro.

Entonces supongamos que existe un punto $p$ en $S^1$ talque ${\rm grad}f(p)=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]=\vec{0}$, pero si elegimos la parametrización $\phi:\ ]0,2\pi[\longrightarrow S^1$ dada por $t\longmapsto\left(\begin{array}{c}\cos(t)\\ \\ \sin(t)\end{array}\right)$, entonces tenemos una función $g=f\circ\phi$ para la cual, la regla de la cadena implica que $g'=f'(\phi)\phi'$ satisface

$\frac{d g}{dt}|_{t_0}=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]\left(\begin{array}{c}-\sin\\ \\ \cos\end{array}\right)_{|_{t_0}}$

i.e.

$\frac{d g(t_0)}{dt}=-\frac{\partial f(p)}{\partial x}\sin(t_0)+\frac{\partial f(p)}{\partial y}\cos(t_0)$

entonces si ${\rm grad}f(p)=\vec{0}$ tendremos $\frac{d g(t_0)}{dt}=0$, en otras palabras $g$ tiene puntos críticos en $t_0$ y en $t_0+2\pi$.

Pero además $g(t_0)=f\circ\phi(t_0)=f(p)$ tanto como

$g(t_0+2\pi)=f\circ\phi(t_0+2\pi)=f\circ\phi(t_0)=f(p)$

es decir $g(t_0)=g(t_0+2\pi)$ y entonces –por el teorema de Rolle– existe $t_1$ en el intervalo abierto $]t_0,t_0+2\pi[$ talque $\frac{d g(t_1)}{dt}=0$.

Pero si nos restringimos a $S^1\setminus\{p\}$ entonces $f=g\circ\phi^{-1}$,
y así (también por la regla de la cadena) tenemos ${\rm grad}f=\frac{dg}{dt}\ {\rm grad}\ \phi^{-1}$ i.e.

$\left[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right]=\frac{dg}{dt}\left[\frac{\partial\phi^{-1}}{\partial x},\frac{\partial\phi^{-1}}{\partial y}\right]$

que evaluando en $t_1$ implica

$\frac{\partial f(q)}{\partial x}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial x}=0$

tanto como

$\frac{\partial f(q)}{\partial y}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial y}=0$

por lo tanto ${\rm grad}f(q)=\vec{0}$, donde $q\neq p$ $\Box$

critical points of functions on the circle