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?

2 Comments

Filed under algebra, free group, group theory, mathematics

2 responses to “is it enough…

  1. No:
    Presuming
    The period of A is 4 and B is 6.
    And AB != BA
    I am pretty sure that I can take an A from the dual cover of GL(2,z3); that is det(A)= -1 and still satisfy the assumptions, but not SL(2,z3).
    Ray

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