Question: We have to determine if there exists a homomorphism from $(\mathbb{Z}_6,+)$ onto $(\mathrm{Z}_3,+)$.

My efforts: Let $\phi$ be an onto homomorphism. Since $\phi$ is surjective, then by the first isomorphism theorem, $\mathbb{Z}_6/\ker\phi \cong \mathrm{Im}(\phi)=\mathbb{Z}_3$. What can I say after this?

Added: Can we say? $\mathbb{Z}_6/\ker\phi \cong \mathrm{Im}(\phi)=\mathbb{Z}_3\implies \left|\mathbb{Z}_6\right|=|\ker\phi||\mathbb{Z}_3|$. Contrapositively, $|\ker\phi||\mathbb{Z}_3|\neq\left|\mathbb{Z}_6\right|\implies $ $\phi$ is not surjective.

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    $\begingroup$ Seems like you're assuming what you're trying to prove, unless $\phi$ is not the onto homomorphism in question. It's not at all clear what you're trying to say in your attempt, regardless. $\endgroup$ – Eevee Trainer May 24 at 2:10
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    $\begingroup$ The mapping $[x]_6\mapsto [x]_3$ is a surjective group homomorphism, where $\left[x\right]_n$ denotes a generic element in $\mathbb{Z}_n$. Just verify that it is well defined and everything is done :) $\endgroup$ – weirdo May 24 at 2:14

Define $\phi: \Bbb Z_6\to \Bbb Z_3$ to be the canonical submersion, where $H=\{0,3\}$ is the kernel.

Note that $H$ forms a subgroup, which is of course normal.

This is the same map @weirdo is talking about.


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