Homomorphism $\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z}$

Question

I think this is just an abuse of notation, but I want to be sure.

Consider the group homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}/3\mathbb{Z}$, where $\mathbb{Z}$ is a group under integer addition and $\mathbb{Z}/3\mathbb{Z}$, and define $\phi(x) = x \text{ (mod $3$)}$.

The notation $x \text{ (mod $3$)}$ is unfamilar to me. I know that what this function is trying to convey is "take the remainder of $x$ when dividing by $3$," but I have never seen it written this way. The proof that this is in fact a homomorphism looks similar: \begin{align*} \phi(x + y) = (x+y) \text{ (mod $3$)} = x \text{ (mod $3$)} + y \text{ (mod $3$)}. \end{align*} Not only does it abuse notation, but it's using the fact that in $\mathbb{Z}_3$, we have $\overline{a + b} = \overline{a} + \overline{b}$, and is therefore treating $x$ and $y$ not as integers (elements of an equivalence class) but representatives of the class itself. Then again, perhaps we're mapping into an equivalence class, so the mapping can be treated as $x \mapsto [x]$.

My question is, am I interpreting this correctly? Is this an abuse of notation? Is there possibly a better way to write this?

Answer 1

Computer scientists and software engineers tend to use the phrase "$m \text{ mod } n$" for what mathematicians might call "the least residue of $m$ mod $n$." Here, to be really formal, "least residue" is the (set-theoretic only, not a homomorphism) section of the surjection $$ \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} $$ which takes the equivalence class $[a]$ to the least non-negative integer $b$ which is equivalent to $a$ mod $n$.

This abuse of notation is rarely harmful, although it does cause people to confuse each other; for instance, is it true that "2 mod 4 + 3 mod 4 equals 5 mod 4"?

Answer 2

So first of all if $x\in\mathbb{Z}$ and $n\in\mathbb{Z}, n>1$ then "$x\text{ mod }n$" is, as you've correctly noted, the reminder of $x$ when divided by $n$.

Now $\mathbb{Z}_n$ can be (and often is) defined literally as $\{0,1,\ldots,n-2,n-1\}$, i.e. the set of all integers between $0$ and $n-1$. Next we define

$$x\oplus y:=(x+y)\text{ mod n}$$

where "$x+y$" is the standard integer addition. This is a well defined binary operation on $\mathbb{Z}_n$ and it turns $(\mathbb{Z}_n,\oplus)$ into an abelian group. And in fact there's a group epimorphism

$$\mathbb{Z}\to \mathbb{Z}_n$$ $$x\mapsto x\text{ mod }n$$

It can be shown that the kernel of that epimorphism is $n\mathbb{Z}$ and so by the first isomorphism theorem our $\mathbb{Z}_n$ group is in fact isomorphic to $\mathbb{Z}/n\mathbb{Z}$. Therefore we can sort of replace "$x\text{ mod }n$" with "$x+n\mathbb{Z}$" and vice versa.

And so many authors use these symbols interchangeably. Which, I agree, can lead to confusion.