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 2021-10-23, 21:49 #1 bhelmes     Mar 2016 32×41 Posts 2*2 Matrix with determinant 1 A peaceful night for you, do I see it right that 2*2 matrix with determinant 1 build a subgroup in linear algebra and that calculation in the group is better for factoring than calculation in all 2*2 matrix ? Do I see right that the matrix (2 0) (0 -1) has the determinant -2 and that an exponentation with 2*p of Mp mod Mp will result in a matrix with determinant 1 Is it rigth that there are 3 pauli-matrixs https://en.wikipedia.org/wiki/Pauli_matrices with determinant 1 and that a parallel multiplication of a start matrix with determinant 1 and a pauli matrix will speed up a possible factorisation ? Would be nice to get a reply. P.S. I do not know why or how, but I am sure we will find the next Mp before next christmas. Last fiddled with by bhelmes on 2021-10-23 at 21:51
 2021-10-24, 07:45 #2 Nick     Dec 2012 The Netherlands 33368 Posts Over any commutative ring R, the 2x2 matrices with determinant 1 form a subgroup $$SL_2(R)$$ of the unit group of the ring of all 2x2 matrices. The Pauli matrices have determinant -1, however. It is important to decide carefully which commutative ring R you are working over: the integers, the Gaussian integers, the complex numbers, etc. Some of these can be constructed using matrices: for example, matrices of the form $\left(\begin{array}{rr}x & -y \\ y & x\end{array}\right)$ over the real numbers form a copy of the complex numbers.
 2021-10-24, 23:21 #3 Dr Sardonicus     Feb 2017 Nowhere 22·5·257 Posts Let's see. det(Mk) = (det(M))k. So if det(M) = -2, det(M2p) = (-2)2p = 22p == 1 (mod 2p - 1). Check. I don't see any relation to factorization.

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