Show that f: $4\mathbb{Z} \to \mathbb{Z}_2$, $f(x)=[x/4]_2$, is a group homomorphism.
I have been trying without much luck to prove this. I have considered individual cases, but noticed the following:
Let $ a,b \in 4\mathbb{Z}$ where 8 divides neither a nor b. Then $f(ab)=[0]_2$, but $f(a)f(b)=[1]_2 [1]_2 = [1]_2$
So $f(ab)\neq f(a)f(b)$
Am I making a mistake here, or did I disprove what I am supposed to prove? Any help or insight would be great, thanks!