How do I find all group homomorphisms from $\mathbb{Z_2} \times \mathbb{Z_2}$ to $\mathbb{Z_6}$?
$\mathbb{Z_2} \times \mathbb{Z_2}$ $= \{ (0,0), (0,1),(1,0),(1,1) \}$
and
$\mathbb{Z_6} = \{0, 1, 2, 3, 4, 5 \}$.
In $\mathbb{Z_2} \times \mathbb{Z_2}$ : $|(0,0)| = 1, |(0,1)| = 2, |(1,0)| = 2,$ and $|(1,1)|=2.$
In $\mathbb{Z_6}$ : $|0| = 1, |1| = 6, |2| = 3, |3| = 2, |4| = 3,$ and $|5| = 6.$
In the case of an isomorphism, elements of a group can only be mapped to elements of the other group of the same order. I understand that in a homomorphism the same rule does not apply.
What are the options for mapping elements of $\mathbb{Z_2} \times \mathbb{Z_2}$ to elements in $\mathbb{Z_6}$? Can elements in the domain be mapped to elements in the codomain that divide such an element?
The trivial homomorphism is just the map that takes $(0,0)$ to $0$.
I also understand that $f(a*b)= f(a) \bullet f(b)$, i.e., an operation preserving map.
Thank you for your time and help.
Here's a useful tool: Let $G$ and $G'$ be groups, let $f \colon G \to G'$ be a group homomorphism. If $x \in G$ and $\vert x \vert = n$, then $$f(x)^n = f(x^n) = f(e) = e,$$ and so $\vert f(x) \vert$ must divide $n$. Using this will cut down significantly in your search for homomorphisms.
Since every nonidentity element of $\mathbb{Z}_2 \times \mathbb{Z}_2$ has order $2$, they must all get sent to an element in $\mathbb{Z}_6$ with order $1$ or $2$.