Why are the elements of the Cyclic Group with Generator $\langle 3 \rangle$ in $\mathbb{Z_{15}}$ {0, 3, 6, 9, 12} and not {1, 3, 9} I'm going through Charles Pinter's A Book of Abstract Algebra which was recommended as an simple introduction.
In the Chapter about Cyclic Groups he mentions that the cyclic subgroup generated by $\langle 3 \rangle$ in $\mathbb{Z_{15}}$ is $\left \{0, 3, 6, 9, 12\right \}$.
I thought I understood that a cyclic group is all powers of its generator? 
Which would lead me to $\mathrm{3}^0$, $\mathrm{3}^1$, $\mathrm{3}^2$
I think I am misunderstanding something badly here.
Could someone please explain to me?
Thank you!
 A: "Power" in the context of groups means repeated application of the group operation to a single element. In general, one usually denotes the group operation multiplicatively, hence the word "power" makes sense. However, in the group of $\Bbb Z_{15}$, there is no multiplication. There is only addition. Addition is the group operation, and the element that would, in the general framework, be denoted as $3^4$ is actually $3+3+3+3 = 12$.
Many people like to have an entirely parallel vocabulary for dealing with additive groups. In that case, they would say that "The cyclic group generated by $3$ consists of all multiples of $3$", and instead of writing $3^4$, they would write $4\cdot 3$, where $4$ is still an integer, and $3$ is still an element of $\Bbb Z_{15}$ (it's harder to see the difference between $4$ and $3$ this way, but it's there).
P.S. Even if we were to consider the "actual" multiplicative powers of $3$ in $\Bbb Z_{15}$, you missed two: $3^3 = 27= 12$ and $3^4 = 81 = 6$.
A: Hint  :  
Here the operation is addition, not multiplication. $$a^5 = a+a+a+a+a$$
