Pascal’s Rule and double induction

If S is a set with \#S=n and T\subset S such that \#T=k then there are {n\choose k} different of them.

Follow the link pascalbernoulli for a double induction proof.

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primer parcial, examen de multi

aplicado ayer √


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three jordan two-by-two matrices and exponential


How many three-by-three forms are there?

: )

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otro lema estratégico

La siguiente lámina establece un lema vital para una demostración “más contemporánea” del Teorema de Sylow 1.


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made in México

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?Nsub3CayleyGcC

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a poem, a strategic lemma


\cdot  p\in{\Bbb{P}}

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

\cdot \forall H<G

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


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


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



double coset counting formula



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wedge product example

When bivectors are defined by


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




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


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


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