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 $(12,16) \neq 1$ .
 A: 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 $\ker f=\langle 4\rangle$.
A: $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$  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, $\ker(f) =\langle 4\rangle$
A: 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 $\mbox{ker}f =<4>.$
