Defining a homomorphism given a mapping and terminology/notation associated?

Question

So the problem is:

"Define a homomorphism $f: (\mathbb{Z}_6, +_6) \ \xrightarrow{onto} (\mathbb{Z}_3, +_3)$.

Explicitly tell me how f is defined: Show f is a function, show f is a homomorphism. "

So the wording confuses me a bit, f is just some mapping so it's telling me to FIND a homomorphism that has that same domain and codomain, correct? If so, I made it a piecewise function

$ f(x) = \left\{\begin{aligned} &[0] &&: x=[0],[3] \\ &[1] &&: x=[1],[4] \\ &[2] &&: x=[2],[5] \\ \end{aligned} \right. $

where $x \ \epsilon \ (\mathbb{Z}_6)$.

Should I put $x \ \epsilon \ (\mathbb{Z}_6, +_6)$ here instead? I feel like that doesn't really workout but I could very easily be wrong, $x \ \epsilon \ (\mathbb{Z}_6)$ seems more natural to me.

Now, would it be better to use $f(x) = [1] : x= [3k+1],$ such that $k = 0, 1, 2, ...$ instead of the x = [1],[4] even though the set is small enough that it's easy to described all possible values of x I'm discussing(and doing this for [0] and [2] as well of course).

Supposing I already showed f is a function, and I go to show it's a homomorphism, is it okay to go about it by just showing $f([x_1] +_6 [x_2]) = f(x_1) +_3 f(x_2)$, where those $x_1, x_2$ values are all the combinations of x that I've listed? Or should I go about it more generally using x=3k, x=3k+1, x=3k+2 (shorthanding a bit here) as described above?

Lastly, for the final conclusion, would the correct form of wording be:

$\therefore f(x)$ is a homomorphism $f: (\mathbb{Z}_6, +_6) \ \xrightarrow{onto} (\mathbb{Z}_3, +_3)$ ?

Or do I just say something like

$\therefore f$ is a homomorphism? All help would be appreciated, just trying to get the proper conventions and understanding down!

Answer 1

The definition you gave is the good one and indeed, in order to show that this is a group morphism you should check the group morphism property for each combination of values $x_1$ and $x_2$. Whereas it is "obvious" that your function is indeed a group morphism, in order to prove it rigorously you should go through a case by case analysis which is not very efficient.

Furthermore, you should not say "$f(x)$ is a homomorphism" you could say "$x\mapsto f(x)$ is a homomorphism". Your second formulation is correct.

If you want to be a little more algebraic you could use the universal property of quotient groups. Indeed $\mathbb{Z}/6$ is $\mathbb{Z}/(6\mathbb{Z})$.

In general, if you are given $G,H$ two groups and $N$ a normal subgroup of $G$. If you want to construct a surjective group morphism $f:G/N\rightarrow H$ it suffices to construct a surjective group morphism $g:G\rightarrow H$ such that $Ker(g)\geq N$. Indeed the universal property of quotient groups tells you that $g$ will factor through the quotient $G/N$ into a function $f:G/N\rightarrow H$ with the same image group.