Vrftjkry

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?"

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

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 $varphi(p^4)$

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