short exact sequence and center


Let us prove:

Let 1\to A\stackrel{f}\to B\stackrel{g}\to B/A\to 1 be a short exact sequence, if the center Z(B/A)=1  then Z(B)<A

Proof:  When x\in Z(B) then g(x)\in Z(B/A), so g(x)=1.

Therefore x\in\ker (g)={\rm im}(f)=A

\Box

 

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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 √

1erexamAlMu

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three jordan two-by-two matrices and exponential


temareas?

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.

truccu

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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


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

\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}<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

\Box

DCCF01

double coset counting formula

 

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