maybe, for the presentation $\langle a,b,c\mid a^2=1, b^2=1 , c^2=1\rangle$, is this the its Cayley’s graph?

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## a poem, a strategic lemma

Lemma

$\cdot$  $p\in{\Bbb{P}}$

$\cdot$ $|G|=p^mr$ , $p\nmid r$

$\cdot$ $\forall H

$\cdot$ $\forall Q\in {\rm p\!-\!SS}_G$

$\Longrightarrow$

$\bullet$ $\exists g\in G$ such that $H\cap gQg^{-1}\in {\rm p\!-\!SS}_H$.

Proof:

By employing the Double Coset Counting Formula we have $|G|=\sum_a\frac{|H|\ |Q|}{|H\cap aQa^{-1}|}$, and since $p\nmid [G:Q]$ then $\exists b\in\{a\}$ such that $p\nmid [H:H\cap bQb^{-1}]$.

But $H\cap bQb^{-1} so $|H\cap bQb^{-1}|=p^l\ ,\ \exists l\in{\Bbb{N}}$, hence, having $p\nmid\frac{|H|}{p^l}$, this implies that $|H|=p^l\alpha$, where $p\nmid\alpha\ ,\ \exists \alpha\in{\Bbb{N}}$.

Then, for $H\cap bQb^{-1} with $|H\cap bQb^{-1}|=p^l$ we deduce it is ${\rm p\!-\!SS}_H$

$\Box$

double coset counting formula

Filed under algebra, group theory, math, word algebra

## wedge product example

When bivectors are defined by

$\beta^i\wedge\beta^j:=\beta^i\otimes\beta^j-\beta^j\otimes\beta^i$,

so, for two generic covectors

$\theta=a\beta^1+b\beta^2+c\beta^3$ and $\phi=d\beta^1+e\beta^2+f\beta^3$,

we have the bivector

$\theta\wedge\phi=(bf-ce)\beta^2\wedge\beta^3+(cd-af)\beta^3\wedge\beta^1+(ad-be)\beta^1\wedge\beta^2$.

Otherwise,

Cf. this with the data $\left(\begin{array}{c}a\\b\\c\end{array}\right)$ and $\left(\begin{array}{c}d\\e\\f\end{array}\right)$ to construct the famous

$\left(\begin{array}{c}a\\b\\c\end{array}\right)\times\left(\begin{array}{c}d\\e\\f\end{array}\right)=\left(\begin{array}{c}bf-ce\\cd-af\\ad-be\end{array}\right)$

So, nobody should be confused about the uses of the symbol $\wedge$ dans le calcul vectoriel XD

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

este proceso se generaliza

2014/08/21 · 13:48

## not all is watching soccer

indeed

Filed under group theory, math

## Kurosh theorem à la Ribes-Steinberg

The strategic diagram is

to plainly grasp the technique