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wreath functoriality


wreath functoriality

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2014/04/03 · 21:20

heptapenti


heptapenti

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2014/03/12 · 18:55

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

Ribes – Steinberg 2008

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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.

Click here to see the complete report.

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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}|}

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<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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