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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?

Added(updated):
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?

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    $\begingroup$ 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 $\phi(6a)$? $\endgroup$ – Jyrki Lahtonen May 23 at 4:03
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    $\begingroup$ 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)... $\endgroup$ – Jyrki Lahtonen May 23 at 4:04
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    $\begingroup$ The relevant phenomena appear in this question. I think calling this an abstract duplicate of that would be a stretch, though. $\endgroup$ – Jyrki Lahtonen May 23 at 4:09
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    $\begingroup$ Good job with the added material! The step $6b=0\implies 3b=0$ looks funny. Where did that come from? $\endgroup$ – Jyrki Lahtonen May 23 at 4:10
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    $\begingroup$ @JyrkiLahtonen I think $6b=0$ should be $2b=0\implies o(b)\leq 2$. Is it correct $\endgroup$ – MKS May 23 at 4:36
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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.

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  • $\begingroup$ +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 $f(g_1)=f(g_2)$. $\endgroup$ – Noah Schweber May 23 at 4:59
  • $\begingroup$ A fair point. My algebra teacher wouldn't often call it Lagrange's "Theorem," the claim being that it isn't very helpful, isn't hard enough to be a theorem, and would perhaps be better put as you've put it. The whole content of the theorem is simply that all cosets have the same size. $\endgroup$ – Charles Hudgins May 23 at 5:04
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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 $$f: x \mapsto 0\;\; \text{and}\;\;f: x \mapsto 2x$$

Can you conclude now?

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