# Determine non-trivial homomorhisms $\phi:\mathbb{Z}_{9}\to \mathbb{Z}_{12}$ and $\phi:\mathbb{Z}_{40}\to \mathbb{Z}_{28}$

Determine non-trivial homomorphisms in each of the following cases:

$$(1)~ \phi:\mathbb{Z}_{15}\to \mathbb{Z}_{45},$$ $$(2) ~\phi:\mathbb{Z}_{9}\to \mathbb{Z}_{12},$$ $$(3) ~\phi:\mathbb{Z}_{40}\to \mathbb{Z}_{28}.$$

Attempt. (1) (Notation: $$[a]_n:=a+n\mathbb{Z}$$ for $$a\in \mathbb{Z}$$ and $$n \in \mathbb{Z}$$). Let $$\phi_1:\mathbb{Z}_{15}\to \mathbb{Z}_{5}\times \mathbb{Z}_{3}:~[a]_{15}\mapsto ([a]_{5},[a]_3),$$ $$\phi_2:\mathbb{Z}_{5}\times \mathbb{Z}_{3}\to \mathbb{Z}_{5}\times \mathbb{Z}_9:~([a]_{5},[b]_3)\mapsto ([a]_{5},[0]_9),$$ $$\phi_3:\mathbb{Z}_{5}\times \mathbb{Z}_{9}\to \mathbb{Z}_{45}:~([a]_{5},[b]_9)\mapsto [-a+2b]_{45},$$ where $$\phi_1,\,\phi_3$$ are isomorphisms and $$\phi_2$$ is homomorphism (see the answer of Construct a non trivial homomorphism $\mathbb Z_{14} \to\mathbb Z_{21}$, noting that $$-1\cdot 9+2\cdot 5=1$$). Then $$\phi_3\circ \phi_2 \circ \phi_1:\mathbb{Z}_{15}\to \mathbb{Z}_{45}:[a]_{15}\mapsto [-a]_{45}$$ defines a non-trivial homomorphism, as wanted.

Regarding the second example, since $$\mathbb{Z}_9 \ncong \mathbb{Z}_3\times \mathbb{Z}_3,$$ we can not proceed as above (using the factor $$\mathbb{Z}_3$$). I found the same difficulty also in $$\mathbb{Z}_{40}$$, where $$\mathbb{Z}_{40}\ncong \mathbb{Z}_4\times \mathbb{Z}_{10}$$.

Thanks in advance for the help.

You want group morphisms right, not ring morphisms?
If so, I would approach the problem differently, let say that you want to establish a non trivial group morphism $$\Bbb Z_a \to \Bbb Z_b$$ where $$a,b \in \Bbb N^*$$.

Let $$k \in \Bbb Z$$ such that $$lcm(a,b)=ka$$, and define $$\phi: \Bbb Z \to \Bbb Z_b$$, $$n \mapsto [kn]_b$$.
Since $$ker(\phi)=(b/k) \Bbb Z$$, and $$(b/k) | a$$ then $$a \Bbb Z \subset ker( \phi)$$, and thus the application $$\phi$$ factors into a $$\Bbb Z_a \to \Bbb Z_b$$ map.

It is non trivial as long as $$k \neq b$$ (you will have $$[k]_b \neq 0$$), so you just have to check that $$a$$ and $$b$$ are not coprime i.e. $$gcd(a,b) \neq 1$$.

Remark: $$k| b$$ by definition of the $$lcm$$, hence $$(b/k)$$ is an integer.

So in your cases, $$a$$ and $$b$$ are not coprime, so the following morphisms are non-trivial:

$$(1)$$ take $$\phi: \Bbb Z_{15} \to \Bbb Z_{45}$$, $$[n]_{15} \mapsto [3n]_{45}$$.

$$(2)$$ take $$\phi: \Bbb Z_{9} \to \Bbb Z_{12}$$, $$[n]_{9} \mapsto [4n]_{12}$$.

$$(3)$$ take $$\phi: \Bbb Z_{40} \to \Bbb Z_{28}$$, $$[n]_{40} \mapsto [7n]_{28}$$.

Edit: What about if $$gcd(a,b)=1$$? In that case, there is no non-trivial morphisms from $$\Bbb Z_a \to \Bbb Z_b$$.

Indeed, if $$\phi: \Bbb Z_a \to \Bbb Z_b$$ is a group morphism, then $$ker(\phi)$$ is a subgroup of $$\Bbb Z_a$$ and $$im(\phi)$$ is a subgroup of $$\Bbb Z_b$$.

Lagrange theorem states that $$card(ker(\phi)) | a$$ and $$card(im(\phi)) |b$$.

Moreover by the first theorem of isomorphism, $$a=card(\ker(\phi)) \times card(im(\phi))$$, thus $$card(im(\phi)) | gcd(a,b)=1$$. Hence the morphism is trivial.