Find the set of homomorphisms from $(\mathbb{Z}_6,+)$ into $(\mathbb{Z}_4,+)$.
My solution goes like this :
We consider $f$ to be a homomorphism from $(\mathbb{Z}_6,+)$ into $(\mathbb{Z}_4,+)$. Now, if $a\in\mathbb{Z}_6)$, then we know that, $f(a^n)=nf(a)$ (Here, $a^n=\underbrace{a.a.a...a}_\text{n times}$, where $n\in\mathbb {Z}$ and . is the binary operation associated in $\mathbb{Z}_6$ which is $+$ in this case , so, $a^n=na$). Thus , $f(a^n)=f(na)=nf(a)$. If $a=[1]$, then, $f([n])=nf([1])$. So, we will completely know $f$ if $f([1])$ is known . Now, $f([6])=6f([1])=f([0])$ since, $[6]=[0]$ . Thus, $6f([1])=f([1])^6=f([0])=0$. So, $o(f([1]))|6$ (, where $o(f([1]))$ denotes the order of $f([1])$). Also, $o(f([1]))|o(\mathbb{Z}_4)$, thus, $o(f([1]))|4$. Thus , $o(f([1]))|gcd(6,4)=2$ and hence, $o(f([1]))=1$ or $2$ . If $o(f([1]))=1$, then $f([1])=[0]$ and so, $f$ is a trivial homomorphism and $f$ maps every element of $\mathbb{Z}_6$ to $[0]\in\mathbb{Z}_4$. If $o(f([1]))=2$, then, $f([1])=[2]$. So, we see , $f([1])=[0]$ or $ [2]$. Thus, $f([n])=[0]$ or $f([n])=[2n]$. These are two homomorphisms possible from $(\mathbb{Z}_6,+)$ into $(\mathbb{Z}_4,+)$.
Is the above solution correct? If not, where is it going wrong ?