Determine kernel and image of homomorphism

Question

Let $\phi : \mathbb{Z}_{18} \rightarrow \mathbb{Z}_{12}$ be a homomorphism with $\phi([1]) = [8]$. What are ker$(\phi)$ and im$(\phi)$?

I am stuck on this problem. The problem that I have is that we are not given the homomorphism explicitly. The kernel is

$$ \text{ker}(\phi) = \{ [a] \in \mathbb{Z}_{18} : \phi([a]) = [0] \}.$$

The image is

$$\text{im}(\phi) = \{ \phi([a]) \in \mathbb{Z}_{12} : [a] \in \mathbb{Z}_{18} \}.$$

Answer 1

Hint: If $\phi([1])=[8]$, then $\phi([2])=[8]+[8]=[4]$, $\phi([3])=[4]+[8]=[0]$, and so on. So, you have enough information to know an explicit description of $\phi$.

Answer 2

Since $\Bbb Z_{18}$ is cyclic, and $1$ is a generator, the homomorphism is completely determined by its effect on $1$.

Then $\rm{ker}\phi=\{0,3,6,9,12,15\}\cong\Bbb Z_6$ and $\rm{im}\phi=\{0,4,8\}\cong\Bbb Z_3$.