# Determine whether there is an onto homomorphism from $(\mathbb{Z}_6,+)$ to $(\mathrm{Z}_3,+)$

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.

• 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. – Eevee Trainer May 24 at 2:10
• 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 :) – 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.