How to find homomorphism of direct product? Let $G$ be the direct product $\mathbb{Z}_4 \times \mathbb{Z}_9$ and let $H$ be the direct product $\mathbb{Z}_4 \times\mathbb{Z}_3 \times\mathbb{Z}_3$. Find a non-trivial homomorphism $\phi : G \to H$
We can tell that $G \ncong H$ because $G$ contains an element of order $36$ while $H$ does not. 
So our homomorphism must be a non-bijective function (because if our homomorphism was bijective, we would have that $G \cong H$ which is a contradiction.)
I tried $\phi \left( (a,b) \right) = \left( a, b \mod 3, b+1 \mod 3 \right) $, which for sure isn't bijective.
I need to show that for arbitrary ordered pairs $x,y \in G$, where $x = (a,b)$ and $y=(c,d)$, that $\phi (xy) = \phi(x) \phi(y)$
I was able to show that it works for a few specific values of $x$ and $y$, but I am having trouble seeing if it is true for all $x,y  \in G$.
I also wonder if my $\phi$ is too complicated and if a more "simpler" homomorphism $\gamma$ exists (that isn't the trivial homomorphism).

What I tried:
$$\phi \left( (a,b) *(c,d) \right) = (a +_4 c, \ [b +_9 d] \mod3,\ [b +_9 d] + 1 \mod3 )$$
where $+_n$ is addition modulo $n$.
Where I'm stuck is seeing if $$\left( a, b \mod 3, b+1 \mod 3 \right)*\left( c, d \mod 3,d+1 \mod 3 \right) =(a +_4 c, \ [b +_9 d] \mod3,\ [b +_9 d] + 1 \mod3 ) $$
Obviously the first part of the ordered 3-tuple is equal, it's the other two that I'm not sure of. The arithmetic with mods in it has me a little lost.
Any help would be greatly appreciated.
 A: $\Bbb{Z}_4\times \Bbb{Z}_9$ is cyclic, with $(1,1)$ as a generator. To obtain a homomorphism $f:\Bbb{Z}_4\times \Bbb{Z}_9 \longrightarrow \Bbb{Z}_4\times \Bbb{Z}_3 \times \Bbb{Z}_3$, all we need is to determine possible images of $f(1,1)$. 
Also note that $f(1,1)=f(1,0)+f(0,1)$. So we will aim at finding possibilities for $f(1,0)$ and $f(0,1)$. Since $|(1,0)|=4$, therefore $|f(1,0)|$ should divide $4$. This means $f(1,0)$ can possibly map to an element of order $4$ (or proper divisors of $4$ as well). Let us say it maps to $(1,0,0)$. 
Likewise, $|(0,1)|=9$, therefore $|f(0,1)|$ should divide $9$. This means $f(0,1)$ can possibly map to an element of order $3$, say $(0,1,1)$ (note: instead of $9$ we are using $3$ because $\Bbb{Z}_4\times \Bbb{Z}_3 \times \Bbb{Z}_3$ has no element of order $9$). 
From this we get,
$$f(a,b)=af(1,0)+bf(0,1)=a(1,0,0)+b(0,1,1)=(a,b,b)$$
If we had made a different choice:say $f(1,0)=(3,0,0)$ and $f(0,1)=(0,1,0)$, then we would have obtained:
$$f(a,b)=af(1,0)+bf(0,1)=a(3,0,0)+b(0,1,0)=(3a,b,0)$$ 
A: You want a ring homomorphism, right ?
There is a homomorphism $\varphi : \mathbb{Z}_9 \to \mathbb{Z}_3$, $(x\: \mathrm{mod} \: 9) \to (x \: \mathrm{mod} \: 3)$. Then $\phi (a,b)=(a,\varphi(b),\varphi(b))$ is a non trivial ring homomorphism.
A: You can try with $\varphi(a,b)=(a,0,0)$.
