Construct a non trivial homomorphism from $\mathbb{Z}_2\times\mathbb{Z}_4$ to $\mathbb{Z}_8$

Question

I don't know how to solve this problem since the group $\mathbb{Z}_2\times\mathbb{Z}_4$ is not cyclic. I just know that $\mathbb{Z}_2 \times \mathbb{Z}_4= \{ (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) \}$ and that the orders of these elements are:

$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$

Can anyone help me?

Answer 1

A homomorphism out of $\mathbb{Z}_2 \times \mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.

Answer 2

Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1\times G_2\to G_3$ is of the form $$ (x,y)\mapsto \alpha(x)+\beta(y) $$ where $\alpha\colon G_1\to G_3$ and $\beta\colon G_2\to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $\alpha$ or $\beta$ is not trivial.

Can you find a nontrivial homomorphism $\mathbb{Z}_2\to\mathbb{Z}_8$?