Homomorphism and group element orders

Question

I return to some old questions that left me puzzled in the past but was none the wiser taking a second look.

Question: Prove that no homomorphism from $\mathbb{Z}_{8}\oplus \mathbb{Z}_{2}$ to $\mathbb{Z}_{4}\oplus \mathbb{Z}_{4}$ exists.

The order of the elements in $\mathbb{Z}_{8}\oplus \mathbb{Z}_{2}$ are 1,2,8 and 4. The order of the elements in $\mathbb{Z}_{4}\oplus \mathbb{Z}_{4}$ are 4. Now, homomorphism does not require preservation of group element orders. One of the condition is that the order of an element g under a homomorphism map divides the order of the element g. However, looking a look at some solutions, it is mentioned that because group element orders are not preserved in this case, the map is not isomorphic and therefore no homomorphism. This doesn't come across as very sensible.

Any help is appreciated.

Thanks in advance.

Answer 1

Your question is not correct as Dietrich Burde mentioned in the comment.

For any groups $G,H$, define $\phi:G\rightarrow H$ by $\phi(g)=1$ for every $g\in G$.
Then $\phi$ is a homomorphism from $G$ to $H$.
This is called the trivial homomorphism.

Answer 2

Also what about homomorphism from $$\Bbb Z_8 \times \Bbb Z_2 \to \Bbb Z_4 \times \Bbb Z_4$$ given by $\phi(x,y)=(x \ mod \ 4, \ x \ mod \ 4)$?