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I'm stumped here. I do know that, if there is such an onto homomorphism, then there must be an isomorphism between $\mathbb{Z}_4\times\mathbb{Z}_4$ / Ker$\phi$ and $\mathbb{Z}_8$. Not sure where to go with that.

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If $\phi:G\to H$ is a homomorphism between finite groups, then the order of $\phi(g)$ in $H$ divides the order of $g$ in $G$.

Next, note that $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$ doesn't contain an element of order $8$.

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Idea:

Suppose $\phi$ is an onto homomorphism from $\mathbb{Z}_{4}\times\mathbb{Z}_{4}$ to $\mathbb{Z}_{8}$. Then $\phi(x)=[3]_{8}$ for some $x\in\mathbb{Z}_{4}\times\mathbb{Z}_{4}$. Derive a contradiction from this.

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