Complexity of solving the discrete logarithm problem for the group formed from product of 2 safe primes
What's that group?


Algebraically, it is isomorphic to the group $\mathbb{Z}_{p-1} \times \mathbb{Z}_{q-1} = \mathbb{Z}_{p'} \times \mathbb{Z}_{q'} \times \mathbb{Z}_2 \times \mathbb{Z}_2$.


  Does $\mathbb{Z}_N^*$ guarantee computing discrete logarithm is hard?


It can be shown that it is at least as hard as factoring $N$ (as if you can compute discrete logs, you can factor).

Hence, if factoring $N$ is infeasible, so is computing discrete logs.


  Since $\mathbb{Z}_N^*$ is not a group of a prime order, is there any guarantee $g$ is a generator.


Actually, there is a guarantee that $g$ is not a generator for the entire group; that's because $\mathbb{Z}_N^*$ is not a cyclic group, and hence does not have a generator.

In particular, the subgroup generated by a single value $g$ will never contain both elements $a$ and $b$ where:


$a \bmod p$ is a Quadratic Residue $\bmod p$, and $a \bmod q$ is a Quadratic Nonresidue $\bmod q$
$b \bmod p$ is a Quadratic Nonresidue $\bmod p$, and $b \bmod q$ is a Quadratic Residue $\bmod q$


(Such elements $a$ and $b$ will always exist in $\mathbb{Z}_N^*$)