Find all homomorphisms from $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$ Find all homomorphisms from $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$
Since $\mathbb Z_4$ is cyclic the homomorphisms are completely determined by the image of $1.$
Possibilities: $$1\mapsto(0,0)\\1\mapsto(0,1)\\1\mapsto(1,0)\\1\mapsto(1,1)$$
All four are homomorphisms, which can be found by direct calculation. E.g. if $\phi:1\mapsto(1,0)$
for $x,y\in \mathbb Z_4$:
\begin{align*}
\phi(x+y)&=(x+y \mod 2,0)\\&=((x \mod 2)+_2(y \mod 2),0)\\&=(x \mod 2,0)+(y \mod 2,0)\\&=\phi(x)+\phi(y).
\end{align*}
Am I correct?
 A: 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.
A: This looks fine to me. You've checked all possible images of $1$ and have seen that all four possibilities give a homomorphism and so you have a full and correct list.
A: Let $\phi(1) = a \implies \phi(x) = xa$, for any $x\in \mathbb{Z}_4$.  $ \ \ \ \phi(x + y) = (x+y)a = xa + ya = \phi(x) + \phi(y)$.  
