There never exists any onto homomorphisms from $(\mathbb{Z_6, +})$ to $(\mathbb{Z_4, +})$. Also find all homomorphisms therein. 
Show that there never exist any onto homomorphisms from $(\mathbb{Z_6, +})$ to $(\mathbb{Z_4, +})$. Also find all homomorphisms therein. 

Here orders of the both the groups are not same. I don't know how to go ahead. Please suggest.  
 A: Here's an outline of one approach:
First observe that any homomorphism $\phi:\mathbb{Z}/6\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ is determined by the image of $1$. Moreover, since 
$$ 2\cdot\phi(1)=6\cdot \phi(1)=\phi(6\cdot 1)=\phi(0)=0$$
it follows that either $\phi(1)=0$ or $\phi(1)=2$. In neither case is $\phi$ surjective.
A: Suppose we have a surjective homomorphism $\phi:\mathbb{Z}_6 \rightarrow \mathbb{Z}_4$.  The isomorphism theorems tell us that $\mathbb{Z}_6 / \ker(\phi) \cong \mathbb{Z_4}$.
We also have $\left| \mathbb{Z}_6/\ker(\phi) \right| = |\mathbb{Z}_6|/|\ker(\phi)|$...
As a hint to the second part of the question, the image of a generator determines any homomorphism out of any cyclic group since $\phi(\underbrace{g + g + \cdots + g}_{\text{ k times} }) = \underbrace{ \phi(g) + \phi(g) + \cdots + \phi(g) }_{ \text{k times} } $ for a generator $g$.
A: Basically it is because $4$ is not a divisor of $6$. Assume we have an onto homomorphism $\phi$. We then have that
$$\Bbb{Z}_6/\ker{\phi}\simeq \Bbb{Z}_4$$
In particular they have the same cardinal. Now we have
$$|\Bbb{Z}_6|=|\Bbb{Z}_6/\ker{\phi}||\ker{\phi}|$$
And this yields 
$${|\Bbb{Z}_6|\over |\Bbb{Z}_4|}=|\ker{\phi}|$$
A contradiction
Now for any homomorphism $\phi$ the image of a generator of $\Bbb{Z}_6$ determines the homomorphism. So we have at maximum four homomorphisms
$\phi(1)=0$ corresponds to the zero homomorphism 
$\phi(1)=1$ means $\phi(2)=2$, $\phi(3)=3$, $\phi(4)=0$ and $\phi(5)=1$. So here the kernel is isomorphic to $\Bbb{Z}_2$
$\phi(1)=2$ means $\phi(2)=0$, $\phi(3)=2$, $\phi(4)=0$ and $\phi(5)=2$. So here the kernel is isomorphic to $\Bbb{Z}_3$
$\phi(1)=3$ means $\phi(2)=2$, $\phi(3)=1$, $\phi(4)=0$ and $\phi(5)=3$. So here the kernel is isomorphic to $\Bbb{Z}_2$
A: Note that $\mathbb{Z}_6\cong\mathbb{Z}_2\times \mathbb{Z}_3$ and any element of order $3$ must be sent to zero. Thus that only leaves $\mathbb{Z}_2$, so the map cannot be surjective.
