# 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

### 2 responses to “is it enough…”

1. rrogers31

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

• rrogers31

Well that didn’t work (: