Category Archives: what is mathematics

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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cf. Frobenius forms


cf. Frobenuis forms

different companion matrices

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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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las básicas


estas son las matemáticas antes llamadas “puras”

Image

o no? :D

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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}.

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problems de álgebra moderna uno: I


Para divertirse estos días de trabajo:

https://juanmarqz.files.wordpress.com/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

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à la Riesz


sabiendo que b^i=g^{si}b_s y que \beta^i(b_j)={\delta^i}_j tanto como \langle b^i,b_j\rangle={\delta^i}_j, entonces

\beta(\quad)=\langle b^i,\quad\rangle

o bien

\beta(X)=\langle b^i,X\rangle

para todo X\in V

 

Ahora para un covector arbitrario f\in V^* se tiene:

f(\quad)

f(X)=f(X^sb_s)=X^sf(b_s)   ————–  (A)

v.s.

f(b_s)\beta^s(\quad)

f(b_s)\beta^s(X)=f(b_s)\langle b^s,X\rangle

=f(b_s)\langle b^s,X^tb_t\rangle

=f(b_s)X^t\langle b^s,b_y\rangle

=f(b_s)X^t{\delta^s}_t

=f(b_s)X^s   —————  (B)

/..\!\!\cdot   (A,B)

f(\quad)=f(b_s)\beta^s(\quad)
=f(b_s)\langle b^s,\quad\rangle
=\langle f(b_s)b^s,\quad\rangle

o bien

f(X)=f(b_s)\beta^s(X)
=f(b_s)\langle b^s,X\rangle
=\langle f(b_s)b^s,X\rangle

para todo X\in V.

Es decir el representante del covector fà la Riesz- es:

f(b_s)b^s

esto es una combinación lineal en la base recíproca b^i de V, which represent the basic covectors \beta^i

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