Task: Find the nilpotents in $\mathbb{Z}/\langle n\rangle$. In particular, take $n=12$.
Solution: An element $m$ is nilpotent $\bmod{n}$ iff $n|m^k$. So $nil(R)=\{0,6\}$.
My question about this solution (provided in textbook) may be some rookie mistake. My first intuition before consulting the solution is that I need to find the order of each element under the group operation of addition. I got $nil(\mathbb{Z}/12\mathbb{Z})=\{0,1,2,3,4,6,8,9,10\}$. I got this by trying to compute $a*b\equiv 0 \mod{12},\, a,b,\in \mathbb{Z}/12\mathbb{Z}$.
The given solution makes sense to me except for the part where $\mathbb{Z}/n\mathbb{Z}$ is an additive group and not multiplication. Could someone explain why this is for me please?
Thanks in advance.