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I am currently working on group theory and more specifically on multiplicative groups. Suppose we have $q = 2 s + 1$ and $p = 2 \pi r + 1$ (where $2, s, r, p, q$ and $\pi$ are different prime numbers). Also we define $N$ as $N = p q$ (like RSA). What is the size of multiplicative group $Z^*_N$ and what is the maximal order that an element $g \in Z^*_N$ can have?

Any help would be appreciated.

Math
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user178592
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  • Are the primes of the form $p = 2ab+1, p,a,b$ prime categorised under some category. I know that primes of the form $p = 2a+1, p,a$ prime are called Sophie Germain prime. – SARTHAK GUPTA Jan 04 '20 at 11:49
  • Since the numbers $p,q$ are prime. The size of multiplicative group $Z^*_N$ will be $\phi(pq) = (p-1)(q-1)$ and I think the maximal order of an element in this group will be $(p-1)(q-1)$ – SARTHAK GUPTA Jan 04 '20 at 11:53
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    That's an unusual meaning for $2\pi r$. – Derek Holt Jan 04 '20 at 12:08

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$\mathbb Z_N^\times \cong C_{p-1} \times C_{q-1} \cong C_{d} \times C_{m}$, where $d=\gcd(p-1,q-1)$ and $m=\operatorname{lcm}(p-1,q-1)$. Therefore, the maximum order of an element in $\mathbb Z_N^\times$ is $m=\operatorname{lcm}(p-1,q-1)=2\pi r s$.

lhf
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  • See also https://math.stackexchange.com/questions/2205618/proving-that-mathbbz-m-oplus-mathbbz-n-cong-mathbbz-d-oplus-mathbbz – lhf Jan 04 '20 at 15:26
  • So by this you mean that the size of multiplicative group is $\mathbb Z_N^\times \cong C_{p-1} \times C_{q-1} \cong C_{d} \times C_{m}$ – user178592 Jan 04 '20 at 16:22