# Group given by congruence relation

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 that

It 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 appreciated

Thanks for your help

Why is it a group?

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 $A^{-1} = 1 - k \cdot B \ mod(N)$

So k is not fixed, you should make it clearer in your question.

With this definition, it is the preimage of a subgroup under the reduction mod N:

`SL(2,Z) -> SL(2,Z/NZ)`

. Hence it is finite index in`SL(2,Z)`

and in particular finitely generated. It is also congruence as it contains Gamma(N).Yes sorry for that, you are right, fixed. Great thanks for this point of view, gives me a step further.

Yes , in fact $G \cap \Gamma_0 = \Gamma(N)$