Comment by vdelecroix for Here is a function constructing the groupdef...
I am glad that it worked! I was suggesting something different though (without going through `M`) libgap.Permutation([[1,1],[0,1]], C, libgap.OnRight) The reason why your version works is that there is...
View ArticleComment by oldani for Here is a function constructing the groupdef...
Got it. Jesus here is how it works for me...don't ask me why... def my_congruence_group(B, N): M = MatrixSpace(Zmod(N), 2) print(M) assert (B**2).is_zero() B = M(B) G = SL(2, Zmod(N)) H = G.subgroup([1...
View ArticleComment by oldani for Here is a function constructing the groupdef...
Thanks. I have the 8.9 version which is recent. I have tried here: https://sagecell.sagemath.org/ and I got the same error....strange. Don't worries I don't want to waste your time I will do my...
View ArticleComment by vdelecroix for Here is a function constructing the groupdef...
I guess you are using a too old version of SageMath (I do not remember when this change happend). You could replace `G([1,1,0,1])` by `[[1,1],[0,1]]` (and the same with the other matrices).
View ArticleComment by oldani for Here is a function constructing the groupdef...
Hi, Sorry to disturb you again but I have tried various ways and I get always the same error when I run the above code: Here is an extract of the fullstack just in case : Starts with:...
View ArticleComment by oldani for Here is a function constructing the groupdef...
Wonderful, thank you so much for your time and your great answer. You taught me a lot in this last two hours. I will study that with great pleasure.
View ArticleAnswer by vdelecroix for Hi,Beginner in Sage (I love it!) , I want to ask you...
Here is a function constructing the group def my_congruence_group(B, N): M = MatrixSpace(Zmod(N), 2) B = M(B) assert (B**2).is_zero() G = SL(2, Zmod(N)) H = G.subgroup([1 + k*B for k in range(N)]) C =...
View ArticleComment by oldani for Hi,Beginner in Sage (I love it!) , I want to ask you...
Yes sorry for that, you are right, fixed. Great thanks for this point of view, gives me a step further. Yes , in fact $G \cap \Gamma_0 = \Gamma(N)$
View ArticleComment by vdelecroix for Hi,Beginner in Sage (I love it!) , I want to ask...
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...
View ArticleComment by vdelecroix for Hi,Beginner in Sage (I love it!) , I want to ask...
So k is not fixed, you should make it clearer in your question.
View ArticleComment by oldani for Hi,Beginner in Sage (I love it!) , I want to ask you...
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...
View ArticleGroup 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, $N$...
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