The general way to find all homomorphism $\mathbb Z_n\to G$ for an arbitray abelian group $G$ is the following: Suppose $\phi:\mathbb Z_n\to G$ is a group homomorphism, as you said, it is determined by the image of $1$, so the question really is which choices of $g\in G$ give a homomorphism $\mathbb Z_n\to G$ when picked as the image of $1$?
For homomorphisms $\psi:\mathbb Z\to G$, this is easy: Pick any $g\in G$, let $\psi(x)=xg$ and just check $$\psi(x+y)=(x+y)g=xg+yg=\psi(x)+\psi(y).$$
Now what could go wrong? The subtle thing that happens here is better exposed, when we write $G$ as a multiplicative group. The definition of $\psi$ becomes $\psi(x)=g^x$ and the calculation becomes $$\psi(x+y)=g^{x+y}=g^x g^y =\psi(x)\psi(y).$$
The important fact: $g^x$ is defined for $x\in\mathbb Z$, that's why $\mathbb Z\to G$ is easy.
Going back to $\phi:\mathbb Z_n\to G$ and sticking to multiplicative notation, it is tempting to just choose $\phi(1)=g$ and define $\phi(x)=g^x$. But here you run into trouble: When $x\in\mathbb Z_n$, what is $g^x$ supposed to mean? Remember that $x\in\mathbb Z_n$ really is a coset of $n\mathbb Z$ in $\mathbb Z$, consisting of all elements of the form $x+kn$ with $k\in\mathbb Z$ and that $\mathbb Z_n$ is really just a shorthand for $\mathbb Z/n\mathbb Z$, the set of all cosets equipped with addition.
What you are really doing is defining $\psi:\mathbb Z\to G$ by $\psi(x)=g^x$ (which is fine) and then look at the induced map $\phi:\mathbb Z_n\to G$ with $\phi(x+n\mathbb Z)=g^x$ for all $x\in\mathbb Z$. Now the definition depends on the representative $x$ of the coset $x+n\mathbb Z$. For the map to be even well defined, it has to be independent of the choice of the representative: If $x+n\mathbb Z=y+n\mathbb Z$, so $x-y=kn$ for some $k\in \mathbb Z$, we want $g^x=g^y$, so $1=g^{x-y}=g^{kn}$. Thus, for $\phi$ to be well defined, we need $g^{kn}=1$ for all $k\in\mathbb Z$: The order of $g$ needs to be a divisor of $n$. When this is the case, the same calculation as above reveals that $\phi$ is not only well defined, but a well defined group homomorphism.
Now that we know, that we can pick any $g\in G$ with order diving $n$, the example $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$ is a quick one: All elements of $\mathbb Z_2\oplus \mathbb Z_2$ are of order $1$ or $2$, both are divisors of $4$, so we can pick any of them.
The group $S_3$ is not abelian, so there is not any injective homomorphism $S_3\to \mathbb{Z}_2\oplus \mathbb{Z}_2\oplus\mathbb{Z}_3$. Thus, any such homomorphism has non-trivial kernel. As the kernel is a normal subgroup of $S_3$ then the kernel of such homomorphism must be equal to $H_2$ or $H_3$. If the kernel is $H_3$ then we have the trivial homomorphism. If the kernel is $H_2$, then as every transposition has order 2, then transpositions must be send to the sumand $\mathbb{Z}_2\oplus \mathbb{Z}_2$. Recall that $S_3$ is generated by a transposition $(12)$ and a $3$-cycle so there are exactly 3 homomorphisms:
- The one sending $(12)$ to $(1,0,0)$ and $H_2$ to $(0,0,0)$;
- The homomorphism sending $(12)$ to $(0,1,0)$ and $H_2$ to $(0,0,0)$; and
- The trivial homomorphism.
Finally, using that $(23)=(132)(12)(123)$ and $(13)=(23)(12)(23)$ you can determine explictly all the above homomorphims.
Best Answer
Any such will be determined by $h(1,0)=m$ and $h(0,1)=n$, and vice-versa, since $\Bbb Z\oplus \Bbb Z=\langle (1,0),(0,1)\rangle $.
Thus $\operatorname {hom}(\Bbb Z\oplus\Bbb Z,\Bbb Z)\cong \Bbb Z\oplus \Bbb Z$.
See here.
Note: $\Bbb Z\oplus\Bbb Z$ isn't cyclic, since its homomorphic image $\Bbb Z_2\oplus\Bbb Z_2$ isn't.