Necessary and sufficient property of a group homomorphism

Question

For groups $G$ and $G'$, if $f: G \to G'$ is a homomorphism, we have for all $x,y \in G$, $f(xy) = f(x)f(y)$. It is easily provable that it is necessary that $f$ carries identity to identity and inverses to inverses.

My question is: are either of these conditions sufficient for $f$ to be a homomorphism? Both follow from the statement that $f(xy) = f(x) f(y)$, but does $f(xy) = f(x)f(y)$ follow from either? Or are there cases where, for example, $f(1_G) = 1_{G'}$ but $f$ is not a homomorphism?

I haven't been able to find a convincing answer to this elsewhere.

Answer 1

No, these two conditions are not sufficient to show that $f$ is a group homomorphim. Consider $f:\mathbb{Z}\to\mathbb{Z}$ by $f(x)=x^3$. Then $f(0)=0$ and $f(-x)=-x^3$. You can see here $f(x+y)\neq x+y$ for any non zero $x$ and $y$ and $x\neq -y$.