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Having two prime numbers $p$, $q$ and a relation $N=pq$.

How could I find the order of group $\mathbb{Z}_{N^2}^{*}$?

I know that $ord(\mathbb{Z}_{N^2})=N^2$ and that $\mathbb{Z}_{N^2}^{*}$ is a subgroup of $\mathbb{Z}_{N^2}$.

Also I know that the order of the group is divided by the order of its subgroups, so $ord(\mathbb{Z}_{N^2}^{*})|ord(\mathbb{Z}_{N^2})$.

So, the possible orders could be either $N^2, N, p , q, 1$, but thats where I get stuck: I don't know how to determine which one would be the right order?

Could you give me any hints how to solve this problem?

Edit:

"What does the ∗ mean in this case?"

$\mathbb{Z}_{n}^{*}=\{a \in \mathbb{Z}_{n} : gcd(a, n) = 1 \}$

Andry
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  • What does the $*$ mean in this case? – Melody Nov 18 '18 at 13:55
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    $\mathbb{Z}^_{N^2}$ is not* a subgroup of $\mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition... – Matt B Nov 18 '18 at 13:56
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    The order of $\mathbb{Z}_{n}^$ is $\phi(n)$. so $\phi(p^2q^2)=p(p-1)q*(q-1)$ Here is the wiki link for $\phi(n)$: https://en.wikipedia.org/wiki/Euler%27s_totient_function also:https://math.stackexchange.com/questions/194705/is-there-a-direct-elementary-proof-of-n-sum-kn-phik – user614287 Nov 18 '18 at 13:56
  • Added definition in my post. @Melody – Andry Nov 18 '18 at 14:00
  • Thanks for pointing out that my problem was already based on my wrong assumption @MattB – Andry Nov 18 '18 at 14:17
  • And thanks for providing the answer @mathnoob – Andry Nov 18 '18 at 14:17

1 Answers1

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First of all, $\mathbb{Z}^*_n$ is not a subgroup of $\mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(\mathbb{Z}^*_n)=\varphi(n)$, so you just have to find $\varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have

$$\varphi(p^2q^2)=\varphi(p^2)\varphi(q^2)$$ $$=(p^2-p)(q^2-q)$$ $$=N^2-Np -Nq+N$$

If $p=q$ we have to calculate $\varphi(p^4)$

$$\varphi(p^4)=p^4-p^3 $$ $$=N^2-Np$$

Bruno Andrades
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