## 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)

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

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

$\Box$

Filed under 3-manifold, algebra, group theory, word algebra

## 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.

Filed under math

1erexamAlMu

Filed under math

## three jordan two-by-two matrices and exponential

How many three-by-three forms are there?

: )

Filed under math

## otro lema estratégico

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

Filed under algebra, cucei math, group theory, what is math

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?

1 Comment

## 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