Subject: how to prove $J_{N}$ a cyclic group? $J_{N}$ is the subgroup of $Z*_{N^2}$ contains all the elements whose jacobi symbol with respect to $N$ is 1.
i saw a paragraph said $J_{N}$ is a cyclic group of order 2NN', $J_{N} $ is the subgroup of $Z*_{N^2}$ contains all the elements whose jacobi symbol with respect to $N$ is 1 ($N=pq, p=2p'+1, q=2q'+1, N'=p'q'$), but i don't know why (especially "cyclic")? Will somebody be kind to help me?
Moreover, it's said $J_{N}=G*G_{1}*G_{2}$, where $G$ is the subgroup of $J_{N}$, which contains all the (2N) residues ($|G|=N'$), $G_{1}$ is a group of order N and $G_{2}$ is the group generated by (-1). WHY?
All the above is to study N-residuosity in $Z*_{N^2}$.
Subject: Re: how to prove $J_{N}$ a cyclic group? $J_{N}$ is the subgroup of $Z*_{N^2}$ contains all the elements whose jacobi symbol with respect to $N$ is 1.
Please don't stuff your postings full of dollar signs.
On 4 Jul, 13:14, challengerlee <lihui0...@gmail.com> wrote:
> i saw a paragraph said $J_{N}$ is a cyclic group of order 2NN', $J_{N} > $ is the subgroup of $Z*_{N^2}$ contains all the elements whose > jacobi symbol with respect to $N$ is 1 ($N=pq, p=2p'+1, q=2q'+1, > N'=p'q'$), but i don't know why (especially "cyclic")?
It isn't cyclic in general. How about N = 341 = 11 x 31. Then (Z/N^2 Z)^* has exponent 30N. Therefore J_N has exponent at most 30N, yet it has 150N elements.
