# Tag Archives: group theory

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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## Kurosh theorem à la Ribes-Steinberg

The strategic diagram is

to plainly grasp the technique

Filed under math

## producto semi-directo

diagram chasing the wreath

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:

$\#(HaK)=\frac{|H||K|}{|H\cap aKa^{-1}|}$

and

$\#(G/K)=\sum_a[H;H\cap aKa^{-1}]$

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 and $n_i$ positive integers.

Let $H$ be a subgroup of $|G|$ of index $[G:H]=q_1$.

Then, $H$ is normal.

Proof:

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:

$|G/H|=1+[H:H\cap a_1Ha_1^{-1}]+\cdots+[H:H\cap a_tHa_t^{-1}]$

i.e.

$q_1=1+\frac{|H|}{|H\cap a_1Ha_1^{-1}|}+\cdots+\frac{|H|}{|H\cap a_tHa_t^{-1}|}$

From this, we get $\frac{|H|}{|H\cap a_iHa_i^{-1}|}.

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.

QED.

## 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 $[G:H]=2$.

Filed under cucei math, math

## 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

$\langle A,B,C$:

$A^2=B^3$,

$A^2B=BA^2$,

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

$AC=CA,BC=CB\rangle$

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 $a^2=b^3$?

http://en.wikipedia.org/wiki/File:SL%282,3%29;_Cayley_table.svg

for more on Cayley tables go

Cayley

(10.nov.2011)

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 $\langle\{a^2,b\}\rangle=\langle b\rangle$.