Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f$.

Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f $.

1- $\langle 2\rangle $

2-$\langle 4\rangle $

3-$\langle 7\rangle $

4-$\langle 1\rangle $

I think first of all there is no group homomorphism with $f(1)=12$ because we know that every homomorphism between cyclic groups sends generators to generators . But in this case $12$ is not generator for $(\mathbb{Z}_{16}, +)$ because $(12,16) \neq 1$ .

$f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism.

Since, $1$ is a generator of $\Bbb Z_{28}$, ${\rm Im}(f) =\langle f(1) \rangle=\langle 12\rangle$ cyclic subgroup of $ \Bbb Z_{16}$

$$|{\rm Im}(f)|= {\rm ord}(12)=4$$

Then by first isomorphism theorem, $\Bbb Z_{28} /\ker (f) \cong{\rm Im}(f) $

Hence, $|\ker(f) |=28/4=7$

So, $\ker(f) $ is a cyclic subgroup of $\Bbb Z_{28}$ generated by an element of order $7$.

And, $\Bbb Z_{28} $ an element of order $ 7 $ is $4$.

Hence, $\ker(f) =\langle 4\rangle$

We have

$$\begin{align} \color{red}{12+12}\pmod{16}&=f(1)+f(1)\\ &=f(1+1)\\ &=f(2)\\ &=\color{red}{8}\pmod{16}, \end{align}$$

$$\begin{align} \color{green}{8+12}\pmod{16}&=f(2)+f(1)\\ &=f(2+1)\\ &=f(3)\\ &=\color{green}{4}\pmod{16}, \end{align}$$

and

$$\begin{align} \color{blue}{4+12}\pmod{16}&=f(3)+f(1)\\ &=f(3+1)\\ &=f(4)\\ &=\color{blue}{0}\pmod{16} \end{align}$$

Thus $\ker f=\langle 4\rangle$.

Observe that $f(2) =8\neq 0, f(7)=4\neq 0, f(1)=12\neq 0$ and $f(4)=0$ so the answer is $\mbox{ker}f =<4>.$