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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,

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  • $\begingroup$ You know the kernel, so you know what goes to zero, so, by default, you know what goes to 1. $\endgroup$ – Gerry Myerson Oct 23 '12 at 12:07
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    $\begingroup$ Note that the second part asks for all homomorphisms, not just the surjective ones. So you need to find the non-surjective homomorphisms as well! $\endgroup$ – Derek Holt Oct 23 '12 at 13:21
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If there is another surjective homomorphism other than the sign homomorphism then we would (putting the two of them together) have a surjective homomorphism $S_4\rightarrow Z_2 \times Z_2$ or equivalently we would have a surjective homomorphism $A_4\rightarrow Z_2$. This leads to a subgroup of order 6 in $A_4$. But it is easy to see this is impossible. So the sign homomorphism is the unique non-trivial homomorphism $S_4\rightarrow Z_2$

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The kernel must contain $\frac{24}2=12$ elements and it must contain all even permutations. There's no room for choice.

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