Why does $\mathbb{Z}/k\mathbb{Z}\cong\mathbb{Z}_{k}$ instead of homomorphism?

Question

I want to learn why does $\mathbb{Z}/k\mathbb{Z}$ isomoprhic to $\mathbb{Z}_{k}$ not homomorphic. For example let's define a map $$\phi:\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}_{2}$$ by $\phi([0])=0$ and $\phi([1])=1$ where [.] is an equivalence class. In text books $\phi$ is said to be a isomorphism.

However I cannot see how $\phi$ is an injective mapping. Because it maps entire class to one element. Doesn't it mean that it maps every element in a class to one element. So how is it injective? Or do we count a class as one element. In that case I can't see why we do it like that.

Answer 1

"Because it maps entire class to one element." In $\Bbb Z/2\Bbb Z$, an entire class is one element. So a function that sends each class to its own element is injective, and if it is also surjective and a homomorphism, then it is an isomorphism.

The map $\phi: \Bbb Z/2\Bbb Z\to \Bbb Z_2$ does come from a map $\phi':\Bbb Z\to \Bbb Z_2$ which is definitely not injective, since $\phi'(0) = \phi'(2)$, but $0\neq 2$. However, for $\phi$, we have $[0] = [2]$, so $\phi([0]) = \phi([2])$ is not a hinder for injectivity.

Answer 2

It is injective because:

• there are only two classes: $[0]$ and $[1]$;
• each class is mapped to a different element.