Suppose $\phi : S_4 \rightarrow \mathbb Z_2$ is a surjective homomorphism. Find $\ker\phi$. Determine all homomorphisms from $S_4$ to $\mathbb Z_2$.
My solution: since $\phi$ is surjective, then by the first isomorphism theorem, $S_4/\ker\phi \cong \mathbb Z_2$. Therefore, by computation, $\ker\phi = A_4$.
Now, for the second part, I have some trouble since they ask for all the possible homomorphisms. I know that the map that takes even permutations to $0$ and odd permutations to $1$ is a homomorphism. But, my question is: How to make sure this is the only one?
thanks,
Best Answer
The kernel must contain $\frac{24}2=12$ elements and it must contain all even permutations. There's no room for choice.