I need to find, and construct, a surjective homomorphism from $A_4$ to $\mathbb{Z}/3\mathbb{Z}$. I have shown that such a homomorphism exists by finding a normal subgroup $H = \{e, (12)(34),(13)(24),(14)(23)\}$ of $A_4$ of order 4 which is the kernel of the homomorphism I am trying to find.
I've constructed the following function $\psi : A_4 \to \mathbb{Z}/3\mathbb{Z}$, $$ \psi(H) = \{\bar{0}\}, \; \psi((123)) = \bar{1}, \; \psi((132)) = \bar{2}, \; \psi((124)) = \bar{2}, \; \psi((142)) = \bar{1}, \; \psi((134)) = \bar{1}, \; \psi((143)) = \bar{2}, \; \text{and} \; \psi((234)) = \bar{2}, \; \psi((243)) = \bar{1}. $$ I need to show that $\psi$ is a surjective homomorphism. By construction it is surjective, so all I have to do is show that it is a homomorphism.
How would I do this?
I assume I'd need to show that $\psi(a + b) = \psi(a) + \psi(b), \; \psi(ab) = \psi(a)\psi(b)$ for all $a,b \in A_4$, but I'm not sure how to do this for permutations.