Given the finite cyclic, additive group (G, +), with |G| = n and generator = g, what are the computations and exchanged messages for Diffie-Hellman?

What I tried myself:

• Alice chooses a private $a$ and sends $p(|G|)$ and $g$ (generator) to Bob.
• Alice calculates $A = a\cdot g \mod p (|G|)$ and sends it to Bob.
• Bob chooses a private $b$ and calculates $B = b \cdot g \mod p (|G|)$ and sends it back to Alice.
• Alice calculates $a \cdot B \mod p (|G|)$ which is the shared key.
• Alice calculates $b \cdot A \mod p (|G|)$ which is the shared key.

Is this way of thinking correct? Because I'm not sure because the group is additive.

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In the beginning you write that the oder of $G$ is $n$ and then you switch to $p$ (or to be precise a strange notation $\pmod{p~|G|}$ which does not make sense)? Furthermore, your write-up does not use the notation of additive groups (but multiplicative ones). – DrLecter Dec 17 '13 at 14:39
How should I change it to use the the additive groups then? – Matthias_164 Dec 17 '13 at 14:42
Additive vs. multiplicative is only a difference in notation. Replace multiplication with addition and exponentiation with scalar multiplication. – CodesInChaos Dec 17 '13 at 14:44
But if you use addition modulo a prime, that's a bad idea, since the discrete logarithm problem in that group is easy. – CodesInChaos Dec 17 '13 at 14:44
@Matthias_164: Is $G$ just an arbitrary group or the integers modulo $N$ (in which case it's probably simpler to rewrite in that notation) – figlesquidge Dec 17 '13 at 14:46

Given the additive group $(G, +)$ with $|G| = p$ and generator $P$, what are the computations and exchanged messages for Diffie-Hellman?
I use order $p$ and assuming $p$ to be prime and the generator as $P$ (as it is used in context of elliptic curve groups - since you need additive groups where the DLP and the CDHP are hard - which is not the case for $\mathbb{Z}_n$).
• Alice chooses random $a\in \mathbb{Z}_p^*$ and sends $A=aP$ (this means $P+\ldots+P$, $a$ times).
• Bob chooses $b\in \mathbb{Z}_p^*$ and sends $B=bP$.
• Alice computes $K=aB=(ab)P$
• Bob computes $K=bA=(ba)P=(ab)P$