What I don't understand is why is it easy to solve the discrete logarithm problem in $\mathbb{Z}/p\mathbb{Z}$. Isn't the Diffie Hellman key exchange based on the difficulty of computing discrete logarithms?
In the additive group $\mathbb{Z}/p\mathbb{Z}$, DLOG is equivalent to computing a modular inverse, which is efficient.
I highly suspect the target of the $p$-adic logarithm is this additive group, as
- the slides use different notation for multiplicative groups,
- you are right --- if it were a multiplicative group DLOG wouldn't be easy, and
- logarithms convert multiplication into addition, so the group operation in the codomain of a logarithm should really be addition.
But even assuming that it is easy do solve the DLP in $\mathbb{Z}/p\mathbb{Z}$, how could I get to the solution to the ECDLP assuming I have the solution to the DLP?
In general, given an efficiently computable injective homomorphism $\phi: G\to G'$, if DLOG is easy in $G'$, then it is easy in $G$.
This is because, given $t = g^a$, one can
- compute $\phi(t) = \phi(g^a) = \phi(g)^a$
- compute the discrete logarithm of $\phi(t)$ to recover $a$.
If $\phi$ isn't an injection you might run into some issues/have to do some more work, but the existence of such a homomorphism is good for attackers generically.
I think pairing-based cryptography has more examples of such things if you're interested in other examples, but don't know details myself.
