Does there exist an onto homomorphism from $(\mathbb{Z}_6,+)$ to $(\mathbb{Z}_4,+)$ and why?

We have to determine whether there exists an onto homomorphism from $$(\mathbb{Z}_6,+)$$ to $$(\mathbb{Z}_4,+)$$.

To do so, let us consider a homomorphism $$\phi:\mathbb{Z}_6\to \mathbb{Z}_4$$. Then $$\phi(a+b)=\phi(a)+\phi(b), \forall a,b\in \mathbb{Z}_6$$. I stuck here. What will be the process to show this?

Let $$a\in \mathbb{Z}_6$$ such that $$\phi(a)=b$$. Then $$b\in \mathrm{Im}\,(\phi)$$.
Now, $$6b=6\phi(a)=\phi(6a)=\phi(0)=0\implies 6b=0\implies 2b=0$$. This implies $$o(b)\leq 2$$. This shows that every element in $$\mathrm{Im}\,(\phi)$$ has order atmost $$2$$. Since $$\phi$$ has to be onto, we must have $$\mathrm{Im}\,(\phi)=\mathbb{Z}_4$$, which contradicts that $$(\mathbb{Z}_4,+)$$ is a cyclic group of order $$4$$. Is my approach correct?

• Hint: Assume that $\phi$ is surjective. Therefore there exists $a\in\Bbb{Z}_6$ such that $\phi(a)=1\in\Bbb{Z}_4$. What can you say about $\phi(6a)$? – Jyrki Lahtonen May 23 at 4:03
• Oh, and did you search the site? You are kinda expected to. We have touched this theme many times (may be even exactly this question, but no guarantees about that)... – Jyrki Lahtonen May 23 at 4:04
• The relevant phenomena appear in this question. I think calling this an abstract duplicate of that would be a stretch, though. – Jyrki Lahtonen May 23 at 4:09
• Good job with the added material! The step $6b=0\implies 3b=0$ looks funny. Where did that come from? – Jyrki Lahtonen May 23 at 4:10
• @JyrkiLahtonen I think $6b=0$ should be $2b=0\implies o(b)\leq 2$. Is it correct – MKS May 23 at 4:36

Suppose there was a surjective homomorphism $$\phi : \mathbb{Z}_6 \to \mathbb{Z}_4$$. Then, by the isomorphism theorem, $$\mathbb{Z}_4 \cong \mathbb{Z}_6 / \ker \phi$$. By Lagrange's theorem, $$|\mathbb{Z}_4| |\ker \phi | = |\mathbb{Z}_6|$$. That is, $$4 |\ker \phi| = 6$$. In other words, $$4$$ divides $$6$$, which is absurd.
We conclude that there is no surjective homomorphism $$\phi : \mathbb{Z}_6 \to \mathbb{Z}_4$$.
I should add that there's a general principle at work here. If $$\phi : G \to H$$ is a surjective homomorphism, then $$|H| \mid |G|$$. So, if $$|H| \nmid |G|$$, then there can be no surjective homomorphism from $$G$$ to $$H$$. To prove this, mimic the proof of the particular case above.
• +1. To some readers: Lagrange's theorem may feel mysterious at first due to the use of the word "kernel," but another way to phrase it is: "given any onto group homomorphism $f:G\rightarrow H$, $f$ sends the same number of elements of $G$ to each element of $H$ - that is, all sets of the form $f^{-1}(\{h\})$ have the same size." Equivalently, any homomorphism $f$ (onto or otherwise) partitions the domain $G$ into a bunch of equal-sized pieces, where $g_1,g_2$ are in the same piece iff $f(g_1)=f(g_2)$. – Noah Schweber May 23 at 4:59
Suppose $$f:\Bbb Z_6 \to \Bbb Z_4$$ be a homomorphism. Let $$f(1)=a$$. Then order of $$f(1)$$ divides both $$6$$ and $$4$$. Thus order of $$f(1)$$ is either $$1$$ or $$2$$. Thus $$a=0$$ or $$a=2$$ are possible. Hence, the number of homomorphism is two. Explicitly, these two are $$f: x \mapsto 0\;\; \text{and}\;\;f: x \mapsto 2x$$