I'm wondering if breaking the DLP, that is the basis for ElGamal and DSA, would automatically break RSA. The way I see it, RSA is based on the following three assumtpions, please correct me if I'm wrong.
The hardness of integer factorization. This is obvious, if this would be possible, we could find p and q from N, calculate phi(N) = (p - 1)(q - 1) and then find d from e in polynomial time using the Extended Euclidian alg.
The hardness of finding e'th rooths mod N. An encrypted message is c = m^e mod N, so knowing c and e (which an attacker does), he could find m by calculating the e'th root mod N.
The hardness of finding The Discrete Logarithm mod N (DLP). After encrypting a message, the attacker knows m and c, and since this relation hold: m = c^d mod N, a solution to the discrete logarithm problem would find d.
When reading different sources, it seems to be the case that solving DLP does not in general break RSA. How is this possible? Does it have to do with the fact that the DLP over a prime order cyclic group (that ElGamal and DSA uses), where all integers less than the group order are members of the set, are concidered to be much easier to solve than over a non-prime group (that (Z_pq, *) is)?
Is it perhaps the case that breaking either 2 or 3 automatically gives a solution to 1 (there already exists polynomial time reduction algorithms), such that the statement "RSA is secure as long as integer factorization is hard" is indeed true.
Thank you in advance for clarifying this!