Number of homomorphisms from $\mathbb{Z}_2\times\mathbb{Z}_2$ to $\mathbb{Z}_4$? Task is in the title.
My attempt:

All elements of $\mathbb{Z}_2\times\mathbb{Z}_2$ have order $2$, except $(0,0)$, which has order $1$. This means that the only possible mapping from $\mathbb{Z}_2\times\mathbb{Z}_2$ to $\mathbb{Z}_4$ is to map $\{(1,0),(0,1),(1,1)\}$ to $2$, since $2$ is the only element in $\mathbb{Z}_4$ that has order $2$, and map $(0,0)$ to $0$. However, consider $(1,0)+(1,1)=(0,1)$. With this homomorphism in mind, we get that $2+2=0$ which would be wrong since $0=(0,0)$, but this doesn't respect group operations from $\mathbb{Z}_2\times\mathbb{Z}_2$ since we should obtain $(0,1)$. Hence, there are no homomorphisms between these groups.

Is this correct?
 A: The correct general rule is that when we're looking at a homomorphism $\phi : G \to H$, if an element $g \in G$ has order $k$, then the order of $\phi(g)$ must be some divisor of $k$.
That's because $\underbrace{g + g + \dots + g}_{k \text{ times}} = e_G$, so $\underbrace{\phi(g) + \phi(g) + \dots + \phi(g)}_{k \text{ times}} = \phi(e_G) = e_H$, but on the other hand, it's possible that a smaller multiple of $\phi(G)$ is also equal to $e_H$.
So in your case, you're right to start by observing that the non-identity elements of $\mathbb Z_2 \times \mathbb Z_2$ have order $2$. What this means is that they must map to elements of order $1$ or $2$ in $\mathbb Z_4$: that is, either to $0$ or to $2$.
You've already ruled out the case where all three of them map to $2$. In fact, turning your argument on its head: once we specify what $(1,0)$ and $(0,1)$ map to, that uniquely determines what $(1,1)$ must map to. This means we have at most four choices: $\phi(1,0)$ is either $0$ or $2$, and $\phi(0,1)$ is either $0$ or $2$.
Check that all four of those choices work.
A: Let $C_2\times C_2=\{a_0:=0,a_1,a_2,a_3:=a_1+a_2\}$ and $f$ be such a homomorphism; then, $f(C_2\times C_2)$ is a proper subgroup of $C_4$ (because $C_2\times C_2\ncong C_4$), namely either $f(C_2\times C_2)=\{0\}$ or $f(C_2\times C_2)=\{0,2\}$. The first case corresponds to the trivial homomorphism $a_i\mapsto 0, i=0,1,2,3$; the second to the $7$ "candidate" maps identified by the condition:
$$f(a_0)=0 \space\wedge\space \exists\alpha \in \{1,2,3\}\mid f(a_\alpha)=2 \tag 1$$
But, these latter are constrained by the condition:
$$f(a_3)=f(a_1+a_2)=f(a_1)+f(a_2) \tag 2$$
which rules out:

*

*$(f(a_1),f(a_2),f(a_3))=(2,2,2)$, because $2\ne 2+2=0$;

*$(f(a_1),f(a_2),f(a_3))=(2,0,0)$, because $0\ne 2+0=2$;

*$(f(a_1),f(a_2),f(a_3))=(0,2,0)$, because $0\ne 0+2=2$;

*$(f(a_1),f(a_2),f(a_3))=(0,0,2)$, because $2\ne 0+0=0$.

So, we are left with the $3$ nontrivial maps:
\begin{alignat}{1}
(f(a_0),f(a_1),f(a_2),f(a_3)) &= (0,2,2,0) \\
&= (0,2,0,2) \\
&= (0,0,2,2) \\
\tag 3
\end{alignat}
which indeed all turn out to be homomorphisms.

$($Alternatively, if we are interested just in the number of such homomorphisms, we can note that there are as many nontrivial of them as the (normal) subgroups of order $2$ of $C_2\times C_2$, which are precisely $3$, namely $N_\alpha:=\{0,a_\alpha\}, \space\alpha=1,2,3$.$)$
