# Find all homomorphisms $φ : \mathbb{Z} → \mathbb{Z}$. Which of these homomorphisms are isomorphisms?

I'm currently working in the following excercise:

Find all homomorphisms $$φ : \mathbb{Z} → \mathbb{Z}$$. Which of these homomorphisms are isomorphisms?

If $$\mathbb{Z} = \mathbb{Z}$$, then would be infinite homomorphisms fulfilling the homomorphism property? I know that an isomorphism is a biyective application of $$φ$$ but I'm not sure about my reasoning about homomorphisms by first.

Thanks in advance for any hint or help and for taking the time to read my question.

• What kind of homomorphisms are you talking about? Of rings or of modules? In any case it will help you to write down the definition of a homomorphism. – Severin Schraven May 27 at 16:54
• Hint: $\phi(n) = n\cdot \phi(1)$ – Jakobian May 27 at 16:54
• @SeverinSchraven Given the tags, I guess group homomorphisms. – Wojowu May 27 at 16:54
• @Wojowu Indeed, you are right. Missed that. Still writing down the definition will solve the problem. Especially after Jakobian's comment – Severin Schraven May 27 at 16:57

## 1 Answer

The image of a map $$\varphi:\mathbb{Z}\to \mathbb{Z}$$ is defined by the image $$\varphi(1)$$, because $$1$$ is a generator of $$\mathbb{Z}$$. For example: if $$\varphi(1)=3$$, then $$\varphi(4)=\varphi(1+1+1+1)=\varphi(1)+\varphi(1)+\varphi(1)+\varphi(1)=12$$, so the map becomes multiplication with $$3$$. Thus we have for each $$n$$ a group homomorphism $$\mathbb{Z}\to \mathbb{Z}$$. In order to be a group isomorphism the generator $$1$$ has to be mapped to a generator, which can only be $$1$$ and $$-1$$, so we have $$2$$ isomorphisms. Does this help?