Homomorphism $f: \mathbb{Z}_{12} \longrightarrow \mathbb{Z}_{30}$

Question

Suppose that we want to construct a non-surjective homomorphism $$ f: \mathbb{Z}_{12} \longrightarrow \mathbb{Z}_{30} $$ Since $\mathbb{Z}_{12}$ is cyclic, $f$ is completely determined from the image of $\overline{1}$ (its generator), $f(\overline{1})$. For the homomorphism to be well-defined, $f(\overline{1})$ must equal $\overline{d}$, where $d$ is a common divisor of $12$ and $30$.

If $f(\overline{1})=\overline{1}$, $f$ is surjective. By excluding this case, we're left with the possible $f$s: $$ f(x)=d \cdot x, \quad d\in \{2,3,6\} $$

Answer 1

The one requirement for $f(1)$ which must be fulfilled is $$ 0=f(0)=f(12\cdot 1)=12f(1) $$ Among elements in $\Bbb Z_{30}$, these are exactly the elements which are multiples of $5$.

Answer 2

(In the second i am submitting, there is already a good accepted answer. But since i started one... This is a very structural answer, the intention was to make clear why there is no chance for $1\to 1$...)

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So we are working in the category of finite abelian groups. (Not in the category of rings.)

An alternative way to see the situation is as follows. Using the Chinese Reminder Theorem, we have isomorphisms: $$ \begin{aligned} \Bbb Z_{12} &\cong \Bbb Z_{2^2}\oplus \Bbb Z_3\ , & 1_{12}&\to1_4\oplus 1_3\\ \Bbb Z_{30} &\cong \Bbb Z_{2}\oplus \Bbb Z_3\oplus \Bbb Z_5\ ,& 1_{30}&\to1_2\oplus 1_3\oplus 1_5\\ \end{aligned} $$ Declaring a map from $\Bbb Z_{12}$ to $\Bbb Z_{30}\cong \color{blue}{\Bbb Z_{2}}\oplus \color{green}{\Bbb Z_3}\oplus \color{red}{\Bbb Z_5}$ is the same as declaring three maps,

$(\ \Bbb Z_{12}\to\color{blue}{\Bbb Z_{2}},\ \Bbb Z_{12}\to\color{green}{\Bbb Z_3},\ \Bbb Z_{12}\to\color{red}{\Bbb Z_5}\ )$.

The last map in the triple is the zero map. So surjectivity is excluded. The other two can be chosen to factor through the corresponding pieces of $\Bbb Z_{12}$. So we finally factor as $$ \Bbb Z_{12}\to \Bbb Z_6 \cong \color{blue}{\Bbb Z_{2}}\oplus \color{green}{\Bbb Z_3} \cong \color{blue}{\Bbb Z_{2}}\oplus \color{green}{\Bbb Z_3} \oplus 0 \to \color{blue}{\Bbb Z_{2}}\oplus \color{green}{\Bbb Z_3}\oplus \color{red}{\Bbb Z_5} \cong\Bbb Z_{30} $$ The first map sends $1$ in some element (without restriction) in $\Bbb Z_6$, there are six possibilities.