# Homomorphism question

## Homework Statement

Consider the mapping $\phi \colon (\mathbb{Z} , + ) \to (\mathbb{Z}, *)$ such that $\phi (a) = a+2$. Define * such that $\phi$ is a homomorphism. For $(\mathbb{Z}, *)$, define the identity element.

## The Attempt at a Solution

Well, first I thought I was oversimplifying this problem. I know that since we have a homomorphism, the identity of $(\mathbb{Z} , \plus )$ will map to the identity element of our image. The identity of $(\mathbb{Z} , \plus )$ will be 0, so $\phi (0) = 1 = e^{\prime}$. But that seemed too easy.

So my question, do I have to break out properties of the homomorphism, namely that $\phi (a+ b) = \phi (a) * \phi (b) = (a+2) * (b+2)$ somehow.

I appreciate any help, thanks!

So for z,z', we need to know what z*z' is. Lets presume first that $$\phi$$ is indeed an homomorphism.
We know that $$z=\phi(z-2)$$ and $$z^\prime=\phi(z^\prime-2)$$. Since $$\phi$$ is a homomorphism:
$$z*z^\prime= \phi(z-2)*\phi(z^\prime-2)=\phi(z+z^\prime-4)=z+z^\prime-2$$
So it is natural to define z*z'=z+z'-2. So now you just need to check that * indeed defines a group structure on Z and that $$\phi$$ indeed defines a homomorphism...