Abstract algebra homomorphism question This was a question on my abstract algebra test. I didn't know what to do with it. 

Let $\phi:\mathbb{Z}\rightarrow \mathbb{Z}_8$, such that $\phi(1) = 6$. For what values of $n$ does $\phi(n) = 0$?
This is a homomorphism.

I listed the multiples of $8$, since they are all equal to $0\bmod 8$. That's obviously wrong, though, and I should be ashamed.
We have $\phi(n)=\phi(1+1+\cdots+1)=n\phi(1)=6n$.

To find $\ker \phi$, we must solve $6n \equiv 0 \bmod 8$.

This equation reduces to $3n \equiv 0 \bmod 4$, or $n \equiv 0 \bmod 4$.

Therefore, $\ker \phi = 4\mathbb Z$.
f(1)=6. f(4)=24=0 . 4 is the least in absolute value that maps to 0. 4Z maps to 0.