Addition / Multiplication modulo 13 I am currently studying Diffie-Helman protocol. In my lecture notes I have a generator g=2 and I'm in group $Z_{13}^*$ . There I calculate a remainder cycle from 1 to 12 with $r =g^i \mod 13  \forall i [0,12]$

I do not quite understand the relation between the generator and the group. This is my $Z_{13}^*$: 



How is the generator in the multiplication group any different from $Z_{13}^+$?
$\mathbb{Z}^*_{13}$ is a group with 12 elements, not 13.

A group is defined by a set of elements, and a "law". The law combines two elements and yields a third one within the set. You get a group if the law fulfils some properties (the law is associative, there is a neutral element, each element has an opposite in the group).

$\mathbb{Z}_{13}$ is a group with 13 elements, using "addition modulo 13" as group law. The neutral element is $0$; the opposite of $4$ is $9$ (because $4 + 9 = 13 = 0 \mod 13$).

$\mathbb{Z}^*_{13}$ is a group that uses "multiplication modulo 13" as group law; it contains all the elements that are invertible modulo 13. Since 13 is prime, these are all elements of $\mathbb{Z}_{13}$ except $0$; so you get 12 elements (the integers $1$, $2$,... $12$ modulo $13$). The neutral element is $1$ (multiplying any element by $1$ leaves it unchanged). The opposite (traditionally called "inverse" since the law is called "multiplication") of $4$ is $10$ (because $4\times 10 = 40 = 1 \mod 13$).

A generator for a group is an element $g$ such that applying the law repeatedly on it ultimately yields all the group elements. In $\mathbb{Z}^*_{13}$, $2$ is a generator for the whole group: if you multiply $2$ by $2$ then you get $4$; if you multiply $4$ by $2$ you get $8$; multiply by $2$ again and you get $3$; then $6$, $12$, $11$, $9$, $5$, $10$, $7$, then $1$. You got all 12 elements. On the other hand, $8$ is not a generator for the whole group since it generates only $8$, $12$, $5$ and $1$.