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

Filed under algebra, cucei math, math, tex

13 responses to “SL_2(Z/3Z), aliquis ordo”

1. Reblogged this on ianmarqz and commented:
SL_2(Z_3)

2. Raymond Rogers

I hope the latex comes through. If not I will repost:

No: Counter example

Take the direct product of $a,b$ $A\times B$ where $a\rightarrow(a,\varepsilon_{B}),b\rightarrow(\varepsilon_{A},b)$

$a\cdot b\rightarrow(a,\varepsilon_{B})\cdot(\varepsilon_{A},b)=(a,b)$
and so on. Or the representation

$\left[\begin{array}{cc} e^{i\frac{\pi}{4}} & 0\\ 0 & e^{i\frac{\pi}{6}}\end{array}\right]=\left[\begin{array}{cc} a & 0\\ 0 & b\end{array}\right]$

These all satisfy $a^{2}b=ba^{2}$ but not the problem.

Alternately take the subgroup $a'=a^{2}$

$a'b=ba'$

Then by commutation $f(a',b)=g(a')\cdot h(b)=h(b)\cdot g(a')$

But no relationship between $b$ and $a,a'$ , other than commutation,
can be inferred.

• with the little make up of inserting “latex ” (latex space) after the first \$ this gains life but it makes no sense…

‘cuz if $a=e^{i\pi/4}$ and $b=e^{i\pi/6}$ then $ab=ba$ which is not the case for the matrices $a,b$ above…

• Your right, the representation was oversimplified; a continuation of the direct product that was too simple.
If you are interested I will post a corrected version.
BTW: I “slipped a gear”; the \pi terms should have been 2\pi.

• Perhaps I can easily clarify: Given the assumptions I can construct a counter example where the conclusion is not true.

• Be my guest.
But let me refrain the quest:
is it true that $a^4=e$, $b^6=e$ and $a^2b=ba^2$ together imply $a^2=b^3$?

• No : Counter example

Let: ${a=\left[\begin{array}{cc} e^{i\frac{2\pi}{4}} & 0\\ 0 & 1\end{array}\right],b=\left[\begin{array}{cc} 1 & 0\\ 0 & e^{i\frac{2\pi}{6}}\end{array}\right],e=\left[\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right]}$

Then: ${a^{4}=e,b^{6}=e,a^{2}b=ba^{2}}$

But: ${a^{2}\neq b^{3}}$

• we would like to see a non abelian one

• “we would like to see a non abelian one”

A little more complicated

——————————————

Counter example

Take: ${u=\left[\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right],v=\left[\begin{array}{cc} 1 & 0\\ 0 & -1\end{array}\right]}$

${w=e^{i\frac{2\pi}{4}},x=e^{i\frac{2\pi}{6}}}$

Then: ${w,x}$ satisfy the conditions and ${u,v}$ satisfy.

${u^{2}=I,v^{2}=I,uv\neq vu}$

Let: ${a=wu,b=xv}$

Then: ${a^{2}=w^{2}I=-I}$

${a^{2}b=w^{2}I\cdot b=b\cdot w^{2}I=ba^{2}}$

But:

${b^{3}=x^{3}v^{3}=\left[\begin{array}{cc} x^{3} & 0\\ 0 & -x^{3}\end{array}\right]=\left[\begin{array}{cc} -1 & 0\\ 0 & 1\end{array}\right]}$

${a^{2}=w^{2}I}$

—————————————————–

Rephrasing:

Counter example

${a=\left[\begin{array}{cc} 0 & e^{i\frac{2\pi}{4}}\\ e^{i\frac{2\pi}{4}} & 0\end{array}\right],b=\left[\begin{array}{cc} e^{i\frac{2\pi}{6}} & 0\\ 0 & -e^{i\frac{2\pi}{6}}\end{array}\right]}$

• Since I am a down to earth kind of guy and like examples;
if your interested in some physical models, I can try to fabricate examples/models of the above groups in the form of.:
Combinatorial necklaces
Solids
Optical systems/paths.
No guarantees though.
Ray

3. Hector

si multiplicamos $a^2=b^3$, por la derecha por $a^4$, tenemos
$a^2a^4=b^3a^4 \Rightarrow$
$a^2e=b^2ba^2a^2 \Rightarrow$
$a^2=b^2a^2ba^2 \Rightarrow$
$a^2=b^2a^2a^2b \Rightarrow$
$a^2=b^2a^4b \Rightarrow$
$a^2=b^2eb \Rightarrow$
$a^2=b^2b \Rightarrow$
$a^2=b^3$

• esto es un buen esfuerzo de práctica de latex, pero el argumento no está bien

4. Hector, no puedes empezar con $a^2=b^3$ porque eso se quiere demostrar a partir de suponer $a^4=e$, $b^6=e$ y $a^2b=ba^2$ solamente.