Category Archives: what is mathematics

2014/01/08 · 10:55

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

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

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

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

Tagged as down voting, free groups, math.SE, mathematics.SE, Schreier

2013/12/09 · 20:35

cf. Frobenius forms

cf. Frobenuis forms

different companion matrices

2 Comments

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

Tagged as companion matrices, Frobenius forms, linear algebra 2, Rational forms

2013/10/18 · 08:24

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:

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

and

$\#(G/K)=\sum_a[H;H\cap aKa^{-1}]$

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:

${\rm Orb}_H(aK)=\{haK\}$ which comply the equi-partition
$HaK=aK\sqcup haK\sqcup...\sqcup h_taK$ , so $\#(HaK)=m|K|$ , for some $m\in \mathbb{N}$
${\rm St}_H(aK)=H\cap aKa^{-1}$

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 $q_1<q_2<...<q_t$ 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 $\frac{|H|}{|H\cap a_iHa_i^{-1}|}<q_1$ .

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

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

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

Then $[H:H\cap a_iHa_i^{-1}]=1$ . So $|H|=|H\cap a_iHa_i^{-1}|$ for each $a_i$ .

This implies $H=H\cap a_iHa_i^{-1}$ and so $H=a_iHa_i^{-1}$ for all the posible $a_i$ , hence, $H$ is normal.

QED.

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Filed under algebra, categoría, category theory, fiber bundle, group theory, math, math analysis, mathematics, maths, what is math, what is mathematics

Tagged as algebra, classes, coset, equivalence relations, group theory, partitions, set theory, Sylow's theorems

2013/08/19 · 20:39

las básicas

estas son las matemáticas antes llamadas “puras”

o no? :D

real elementary multilinear algebra

are the common algebraic-techniques territory for today vector algebra and differential geometry.

This means that ancient vectorcalculus that turns into differential forms nowadays, is “super-oversimplified” into a mathematical language to phrase some modern geometricalalgebrotopologicalanalitic-maths.

Real, ‘cuz first you gotta get the ideas over the field $\mathbb{R}$ .

2 Comments

Filed under algebra, calculus on manifolds, category theory, differential geometry, geometry, math analysis, mathematics, multilinear algebra, topology, what is math, what is mathematics

Tagged as algebra, álgebra multilineal, cucei, differential forms, multilinear algebra

2011/10/06 · 14:22

problems de álgebra moderna uno: I

Para divertirse estos días de trabajo:

https://juanmarqz.wordpress.com/wp-content/uploads/2011/10/canonproblms.pdf

Samples:

Sea $f:S\to T$ un map. Sean $A,B\subset S$ entonces

i) ¿ $f(A\cap B)=f(A)\cap f(B)$ ?

Solución: Es fácil la contención $f(A\cap B)\subseteq f(A)\cap f(B)$ . La otra contención $f(A\cap B)\supseteq f(A)\cap f(B)$ no siempre es posible: pues si $t\in f(A)\cap f(B)$ entonces $t\in f(A)$ y $t\in f(B)$ , esto implica que existen $c\in A$ y $d\in B$ tales que $f(c)=f(d)=t$ . Pero esto no implica que exista $a\in A\cap B$ talque $f(a)=t$ …

ii) Sean $X,Y\subset T$ . Demuestra $f^{-1}(X\cup Y)=f^{-1}(X)\cup f^{-1}(Y)$

La función $\phi(n,m)=m+\frac{(n+m)(n+m+1)}{2}$ mapea ${\mathbb{N}}\times{\mathbb{N}}\to{\mathbb{N}}$ . Demuestra que es una biyección. Solución.

Sea $f:S\to T$ un map sobreyectivo. Demuestra que si para cualquiera dos $x,y\in S$ definimos: $x\sim y\ \Longleftrightarrow f(x)=f(y)$ , entonces esto determina una relación de equivalencia en $S$
Demuestra que el map ${\mathbb{R}}\to (-1,1)$ , de los reales al intervalo abierto $x\in{\mathbb{R}}: -1<x<1$ , dando por

$x\mapsto \frac{x}{1+|x|}$

es una biyección. Solución: Aquí algunos detalles pero desordenados: El cero se mapea en el cero. Note que $\frac{x}{1+|x|}$ tiene el mismo signo que $x$ . Note también que $x<1+|x|$ entonces $|\frac{x}{1+|x|}|<1$ y esto implica que el codominio es efectivamente el intervalo abierto $(-1,1)$ o sea este map esta bien definido. Para la inyectividad hay varios casos según se elijan $a,b$ : ambos $a,b>0$ ó $a,b<0$ ó $a<0<b$ ó $a>0>b$ , pero de todos modos si $a\neq b$ y aunque pueda suceder $1+|a|=1+|b|$ tendremos $\frac{a}{1+|a|}\neq \frac{b}{1+|b|}$ . Para la sobreyectividad sea $p\in (-1,1)$ para el cual buscamos $c\in{\mathbb{R}}$ talque $\frac{c}{1+|c|}=p$ . Entonces tenemos que resolver $\frac{c}{1+c}=p$ para $p>0$ o bien $\frac{c}{1-c}=p$ para $p<0$ . En el primer caso $c=\frac{p}{1-p}$ y en el otro $c=\frac{p}{1+p}$ .

$\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)\left(\begin{array}{cc}1&2\\ 0&1\end{array}\right)=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$ , con entradas en ${\mathbb{Z}}_3$
$\left(\begin{array}{cc}2&2\\ 2&1\end{array}\right)\left(\begin{array}{cc}1&2\\ 1&2\end{array}\right)=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$ , también con entradas en ${\mathbb{Z}}_3$