Let $f:\mathbb{Z} \rightarrow \mathbb{Z}_{10}$ is homomorphism. Determine $f(18)$ such that $f(1)=6$.

Question

Let $f:\mathbb{Z} \rightarrow \mathbb{Z}_{10}$ is homomorphism. Find $\ker f$ and $f(18)$ such that $f(1)=6$.

Here I tried.

$$\ker f = \lbrace 10q \;|\; q \in \mathbb{Z} \rbrace.$$

Next, may I assume that $f(n)=\bar{n}, \forall n \in \mathbb{Z}$?

If yes, it should be $f(1)=\bar{6}$, right? Then, $f(2)=f(1)+_{10}f(1)\equiv 2 \pmod {10}$ and so on. Hence, $f(18) \equiv 8 \pmod {10}$.

Thanks for help in advanced.

Answer 1

$f(18)\equiv8\bmod10$ is right, but your kernel is wrong. Since $$f(n)=f(\underbrace{1+1+\cdots+1}_{n})=\underbrace{6+6+\cdots+6}_n\equiv6n\bmod10$$ it follows that $\ker f=\{5a:a\in\mathbb Z\}$. The multiplier is not 10; $f(5)\equiv0\bmod10$.