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?,3%29;_Cayley_table.svg

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.


Filed under algebra, cucei math, math, tex

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

  1. 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}


    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]}


        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}}


        {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]}




        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
        Optical systems/paths.
        No guarantees though.

  2. 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

  3. 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.

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