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Suppose that $N=PQ$, such that $P=2p+1,Q=2q+1$ and $P,Q,p,q$ are all primes. Then the order of $\mathbb{Z}_N^*$ is $4pq$. Then the order of subgroup of $\mathbb{Z}_N^*$ is the factor of $4pq$.

My question is : what's the number of subgroup in $\mathbb{Z}_N^*$, e.g. the number of subgroup with order $4$?

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  • $\begingroup$ Groups of order $4$ are isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. So, you have to check which one of them first. $\endgroup$ Commented May 13, 2014 at 4:31

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If you are looking for the number of subgroups of a cyclic group of order $n$, it is $d(n)$, the number of divisors of $n$. So if $n=4pq$, $p$ and $q$ different odd primes, then $d(n)=3.2.2=12$.

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  • $\begingroup$ The group is not cyclic. $\endgroup$ Commented May 13, 2014 at 7:43
  • $\begingroup$ no, it's not I want, e.g. in the $\mathbb{Z}_N^*$, there is a subgroup $\mathbb{G}$ and $\mathbb{G} \subset \mathbb{Z}_N^*$. Now I suppose the order of $\mathbb{G}$ is $p$. Then, what I want to know is how many different subgroups with order $p$ in $\mathbb{Z}_N^*$. things like that .. $\endgroup$
    – Nax
    Commented May 13, 2014 at 10:30

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