Prove this mapping is a homomorphism

Here is the example:

The mapping from S_n to Z_2 that takes an even permutation to 0 and an odd permutation to 1 is a homomorphism.

-I have to prove that this example is a homomorphism.

I have looked at other homomorphisms and I understand how to prove them. They are pretty easy, such as determinants or derivatives. For some reason this one just eludes me, and I feel that there is a simple answer. i know that i have to prove phi(ab)=phi(a)phi(b) . this is an additive group so i think it would be phi(a+b)=phi(a)+phi(b) right? and i maybe have to do two cases one for even, one for odd? Thanks for help in advance.

Re: Prove this mapping is a homomorphism

no, the group operation in $\displaystyle S_n$ is composition, so what you want to prove is:

$\displaystyle \varphi(\sigma \circ \tau) = \varphi(\sigma) + \varphi(\tau)$, for $\displaystyle \sigma,\tau \in S_n$.

what this boils down to is proving:

even composed with even is even

even composed with odd is odd

odd composed with even is odd

odd composed with odd is even