Tag Archives: group theory

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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Kurosh theorem à la Ribes-Steinberg

The strategic diagram is


to plainly grasp the technique


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producto semi-directo

producto semi-directo

diagram chasing the wreath

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2014/05/03 · 14:50

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}|}


\#(G/K)=\sum_a[H;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\} which comply the equi-partition
  • HaK=aK\sqcup haK\sqcup...\sqcup h_taK, so \#(HaK)=m|K|, for some m\in \mathbb{N}
  • {\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.


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}]


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.



Filed under algebra, categoría, category theory, fiber bundle, group theory, math, math analysis, mathematics, maths, what is math, what is mathematics

a group

For G=\langle a,b\ |\ a^2=e,\ b^2=e\rangle, let us write some elements:

word length : words

0 : e

1 : a,b

2 : ab,\quad ba

3 : aba,\quad bab

4 : abab,\quad baba

5 : ababa,\quad babab

6 : ababab,\quad bababa

Note that for odd-length-words, like aba we have: (aba)(aba)=e,

but for even-length-words, like abab, we have: (abab)(baba)=e.

So, this group has infinite many elements of order two and the even-length words form a subgroup, H, isomorphic to the group of the integers, and its index is [G:H]=2.

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is it enough…

that the presentation given by

\langle A,B\ :\ A^2=B^3,\quad A^2B=BA^2,\quad A^4=e,\quad B^6=e\rangle

determines the group SL_2({\mathbb{Z}}/{3\mathbb{Z}})?

Similar question for

\langle A,B,C:



A^4=B^6= C^2=e,


for the group GL_2({\mathbb{Z}}/{3\mathbb{Z}}).

Other similar problems but less “difficult” are:

  • \langle\varnothing:\varnothing\rangle=\{e\}
  • \langle A\ :\ A^2=e\rangle  for  \mathbb{Z}_2
  • \langle A\ :\ A^3=e\rangle  for  \mathbb{Z}_3
  • \langle A\ :\ A^4=e\rangle  for  \mathbb{Z}_4
  • \langle A,B\ :\ A^2=e, B^2=e, AB=BA\rangle  for  \mathbb{Z}_2\oplus\mathbb{Z}_2
  • \langle A,B\ :\ A^2=e, B^3=e, AB=B^2A\rangle  for  S_3
  • \langle A,B\ :\ A^2=e, B^3=e, AB=BA\rangle  for  \mathbb{Z}_2\oplus\mathbb{Z}_3
  • \langle A:\varnothing\rangle it is \mathbb{Z}
  • \langle A,B:AB=BA\rangle  is \mathbb{Z}\oplus\mathbb{Z}
  • \langle A,B:\varnothing\rangle  is \mathbb{Z}*\mathbb{Z}, the rank two free group
  • \langle A,B,J:(ABA)^4=e,ABA=BAB\rangle for SL_2(\mathbb{Z})
  • \langle A,B,J:(ABA)^4=J^2=e,ABA=BAB, JAJA=JBJB=e \rangle for GL_2(\mathbb{Z})
  • \langle A,B:A^2=B^3=e \rangle for P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3
  • dare altri venti esempi
Qual è il tuo preferito?


Filed under algebra, free group, group theory, mathematics

SL_2(Z/3Z), aliquis ordo

hae sunt matrices (estas son las matrices):

a=\left(\begin{array}{cc}0&1\\ 2&0\end{array}\right) , b=\left(\begin{array}{cc}0&1\\ 2&1\end{array}\right) , ba^3b=\left(\begin{array}{cc}0&1\\ 2&2\end{array}\right)

a^3=\left(\begin{array}{cc}0&2\\ 1&0\end{array}\right) , bab=\left(\begin{array}{cc}0&2\\ 1&1\end{array}\right) , a^2b=\left(\begin{array}{cc}0&2\\ 1&2\end{array}\right)

e=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right) , a^2ba=\left(\begin{array}{cc}1&0\\ 1&1\end{array}\right) , (ba)^2=\left(\begin{array}{cc}1&0\\ 2&1\end{array}\right)

(ab)^2=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right) , b^2ab=\left(\begin{array}{cc}1&1\\ 1&2\end{array}\right) , a^3ba=\left(\begin{array}{cc}1&1\\ 2&0\end{array}\right)

a^3b=\left(\begin{array}{cc}1&2\\ 0&1\end{array}\right) , a^2b^2=\left(\begin{array}{cc}1&2\\ 1&0\end{array}\right) , bab^2=\left(\begin{array}{cc}1&2\\ 2&2\end{array}\right)

a^2=\left(\begin{array}{cc}2&0\\ 0&2\end{array}\right) , ab^2=\left(\begin{array}{cc}2&0\\ 1&2\end{array}\right) , ba=\left(\begin{array}{cc}2&0\\ 2&2\end{array}\right)

ab=\left(\begin{array}{cc}2&1\\ 0&2\end{array}\right) , b^2aba=\left(\begin{array}{cc}2&1\\ 1&1\end{array}\right) , b^2=\left(\begin{array}{cc}2&1\\ 2&0\end{array}\right)

b^2a=\left(\begin{array}{cc}2&2\\ 0&2\end{array}\right) , aba=\left(\begin{array}{cc}2&2\\ 1&0\end{array}\right) abab^2=\left(\begin{array}{cc}2&2\\ 2&1\end{array}\right)

… ante hoc dictionary

Nunc, possibile est quod haec propositio quaerere:

\langle a,b\ :\ a^4=e, b^6=e, a^2b=ba^2\rangle

  • is it possible that: a^2b=ba^2 together with a^4=e and b^6=e they imply a^2=b^3?


for more on Cayley tables go













Usando la teoría básica de grupos podremos construir un subgrupo de 12 elementos usando un subgrupo de orden tres y otro subgrupo de orden cuatro:

Con H=\langle p|p^3=e\rangle y con K=\langle q|q^4=e\rangle armamos HK. Así HK tendrá 12 elementos y pues H\cap K=\{e\}.

Chéquelo con p=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)  y q=\left(\begin{array}{cc}1&1\\ 1&2\end{array}\right).

El que queriamos armar usando \langle\{a^2,b\}\rangle, no funciona para obtener un subgrupo de 12 elementos,  pues porque, sabiendo a^2=b^3, tendremos \langle a^2\rangle<\langle b\rangle y por lo tanto \langle\{a^2,b\}\rangle=\langle b\rangle.


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