Homomorphism and image

Question

I'll consider the function $\phi: \mathbb Z^3 \to \mathbb Z^3$, given by:

$\phi(x,y,z) := (x+5y+3z,2y,7z)$

I have a couple of questions associated with this:

1) This might be a very dumb question, but if I wanted to show, that this is an homomorphism, I'll just follow the standard approach, but am I able to just add the 5y with the 2y etc. or how should I interpret the commas?:

I'll say that x,y,z,a,b,c are all elements of the integers. Then:

$\phi(x,y,z,a,b,c)=\phi(x+a,y+b,z+c)=x+a)+5(y+b)+2(y+b)+3(x+c)+7(x+c)=(x+a)+7(y+b)+10(z+c)=(x+7y+10z)+(a+7b+10c)=\phi(x,y,x)+\phi(a,b,c)$

It is thereby an homomorphism.

So my question here is simply if I'm allowed to do this or how else I should interpret the commas in the function?

2) Secondly, I'll have that the image of $\phi(\mathbb Z^3)$ is just $\mathrm{Im}\ \phi(\mathbb Z^3) = \{\phi(x,y,z) : x,y,z \text{ being elements of }\mathbb Z^3\}$ (right?), but how would I go about finding the number of elements in $\mathbb Z^3/\phi(\mathbb Z^3)$?

3) Last, but not least, I was wondering; What possibilities are there generally for the structure of the image for a homomorphism: $\psi: \mathbb Z^3 \to \mathbb Z^3$?

Any help would be deeply appreciated.

Answer 1

1. The function $\phi$ is a function with three variables, not six, and therefore the expression $\phi(x,y,z,a,b,c))$ makes no sense. What you are supposed to prove is that$$\bigl(\forall(x,y,z),(a,b,c)\in\mathbb Z^3\bigr):\phi\bigl((x,y,z)+(a,b,c)\bigr)=\phi(x,y,z)+\phi(a,b,c).$$
2. Yes, $\operatorname{Im}\phi=\bigl\{\phi(x,y,z)\,|\,(x,y,z)\in\mathbb Z^3\bigr\}$, but the idea is to determine this set explicitely. For instance, if $(a,b,c)\in\operatorname{Im}\phi$, then $7\mid c$. After you have determined $\operatorname{Im}\phi$ explicitely, then you can try to comput $\mathbb Z^3/\operatorname{Im}\phi$.
3. There are several possibilites, such has $\mathbb Z$, $\mathbb Z^2$, $\mathbb Z_2\oplus\mathbb Z$, and so on.