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Mar 28, 2022 at 14:31 comment added François Brunault For reference, what was wrong is that in Shimura's description, the condition $(c,d)=1$ should be replaced by $(c,d,N/d)=1$, see the answers by Max Horn and me.
Apr 7, 2016 at 0:45 history edited Michael Albanese CC BY-SA 3.0
added 2 characters in body
Dec 20, 2011 at 23:08 answer added François Brunault timeline score: 3
Dec 20, 2011 at 22:21 comment added François Brunault The output by Sage is correct, I think it gives you representatives for the left quotient $\Gamma_0(N) \backslash \Gamma$.
Dec 20, 2011 at 14:27 answer added George McNinch timeline score: 1
Dec 20, 2011 at 11:43 answer added Nick timeline score: 0
Dec 19, 2011 at 17:30 comment added Max Horn Yes, for prime $N$, we have that $R=Z/NZ$ is a finite field, in particular all elements are units, and it is well known that the projective line has the coset representatives $(1,1),\ldots,(1,N),(N,1)=(0,1)$.
Dec 19, 2011 at 15:57 answer added Igor Rivin timeline score: 1
Dec 19, 2011 at 15:48 comment added Igor Rivin Does everything work for prime $N?$
Dec 19, 2011 at 15:17 answer added Max Horn timeline score: 3
Dec 19, 2011 at 13:10 comment added Max Horn Actually, my comment (and my now deleted answer) is incorrect. Assuming you mean the group of matrices $\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)$ for which $c$ is divisible by $N$, then of course $1\in\Gamma$ is in the coset represented by $(N,1)$ (and (0,1) is just another rep for that). Moreover, $(1,0)$ and $(1,N)$ also represent the same coset.
Dec 19, 2011 at 12:37 history edited Max Horn CC BY-SA 3.0
Rewrote the matrices to be clearer (I did not understand them at all beforehand)
Dec 19, 2011 at 12:32 comment added Max Horn The missing representatives are (0,1) for the coset of 1, and (1,0).
Dec 19, 2011 at 11:20 history asked Nick CC BY-SA 3.0