Comment by oldani for Hi,Beginner in Sage (I love it!) , I want to ask you this maybe naive question:I have a group $G$ defined by $\lbrace M \in SL_2(Z) \ | \ M = I + kB \ mod(N) \rbrace $, where $I$ is the identity, $N$ a given positive integer, $B$ is a given fixed matrix verifying $B^2 = 0$ and $k$ (not fixed) any integer in { 0,1,2,....,N-1 }. I want to explore conjugacy classes/subgroups of this group $G$ for some specific $B$ and $N$. Is this subgroup finitely generated for some $N$? subgroups of finite index? etc... Do I have a way to do that with SageMath?Remark thatIt is subgroup of a finitely generated group butv that do not imply that it is finitely generated$B^2 = 0$ gives that $I + kB = \Lambda^k \ with \ \Lambda = I + B$. An obvious subgroup is $ H = \lbrace \Lambda^k, \ k \in \mathbb Z \rbrace$Any advise or web pointer would be appreciatedThanks for your help

Hi, Thanks for your interest in my question . Here is the argument Essentially because $B^2 = 0$. Just to be clear, saying that $A=B \ mod(N)$ means that $A - B = N \cdot X$ where $X$ is an integral matrix 1) $I \in G$, take k=0 2) Take $A = I + mB + N\cdot X, B = I + nB + N\cdot Y\in G$ , $X,Y$ being any integer coefficients matrices. Write the product and you will see that (because $B^2=0$) we have $A \cdot B = I + (m+n)\cdot B \ mod(N)$ 3) For the inverse, you see easily that if $A = 1 + k \cdot B \mod(N)$ then $A^{-1} = 1 - k \cdot B \ mod(N)$