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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?

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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.

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  • $\begingroup$ Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$\in \mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$\cdot$2=4 ,so f(0,0)=2$\cdot$4=8=0 in $\mathbb{Z_8}$? I don't know if my thoughts are right. $\endgroup$ – Failousa Jan 4 at 19:43
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    $\begingroup$ $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that. $\endgroup$ – user3482749 Jan 4 at 19:46
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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 $\mathbb{Z}_2\to\mathbb{Z}_8$?

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  • $\begingroup$ I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $\mathbb{Z_2}\to\mathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it? $\endgroup$ – Failousa Jan 4 at 19:58
  • $\begingroup$ @Failousa You can prove it directly. $\endgroup$ – egreg Jan 4 at 20:25

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