# proving homomorphism versus proving group homomorphism

Quoting " Let $\phi : Z \rightarrow Z$ be given by $\phi(n) = 7n$. Prove that $\phi$ is a group homomorphism. Find the kernel and the image of $\phi$."

Whether proving that we have a homomorphism or proving we have a group homomorphism, isn't it the same proof process?

I prove that it is a group homomorphism by showing that: $$\phi(x \circ y) =\phi(x) . \phi(y)$$ ($x,y \in Z$ and group operation $\circ$ and $.$ are both addition).

Any input is much appreciated.

## 1 Answer

No, it is not the same. The integers also form a ring, so we could ask for ring homomorphism or group homomorphism - which are two different things. A group homomorphism would require $\phi(x+y)=\phi(x)+\phi(y)$ for the group $(\mathbb{Z},+)$, and a ring homomorphism for the ring $(\mathbb{Z},+,\cdot)$ in addition that $\phi(x\cdot y)=\phi(x)\cdot \phi(y)$ for all $x,y$.

• Thx for the input. We haven't look at "ring homomorphism" yet. The problem i have is that I am facing exercises just stating "prove that the function is a "homomorphism" and other exercises stating "prove that the function is a group homomorphism". – gegu Apr 15 '17 at 20:24
• A homomorphism always refers to an algebraic structure, i.e., a homomorphism of what? Of a group, a ring, a field, etc.? So speaking of homomorphisms, always assumes that we have said before, what we mean. I suppose in your case it is always groups (because you haven't looked at rings or fields). – Dietrich Burde Apr 15 '17 at 20:25