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I tried to solve an exercise from a previous test asking "find all the homomorphisms from $~S_4~$ to $~Z_4~$ and show that those are all of them"

In the solution they used homo' such that:

$\phi(s)=0~$ if $~s~$ is in $~A_4$

$\phi(s)=2~$ if $~s~$ is not in $~A_4$

and because $~A_4~$ is contained in the Kernell they said that those are all the homo'there are can please someone explain to me why is that true?

Perhaps I didn't understand the solution?

thank you!

nmasanta
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E. Ginzburg
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  • That's not the only homomorphism. You also have the trivial homomorphism, that sends every element to $0$. And don't you mean $\phi(s)=2$ if $s$ is not in $A_4$? Or do you want homomorphisms to $\mathbb{Z}_2$ and not $\mathbb{Z}_4$? Because otherwise, you can take two elements $s_1$ and $s_2$ which are not in $A_4$ and get $\phi(s_1)+\phi(s_2)=2\neq\phi(s_1s_2)$, since no element has image $2$, so it would not be a homomorphism. – Paulo Mourão Jul 12 '19 at 07:30
  • yes, i meant phi(s) = 2, thank you.

    but how do i show that those two are the only one's there are?

     – E. Ginzburg Jul 12 '19 at 07:39

1 Answers1

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Well, first things first, that is not the only homomorphism. You also have the homomorphism that takes every element to $0$.

That justification seems to be using the fact that $A_4$ is the commutator subgroup of $S_4$. You can see a proof of this fact here.

Now, given an arbitrary group $G$, the commutator subgroup is usually denoted $[G,G]$ and you can show that it's a normal subgroup of $G$. The quotient $G/[G,G]$ is called the abelianization of $G$ and it has the following universal property:

Given an abelian group $H$ and a homomorphism $f:G\to H$, there exists a unique homomorphism $g:G/[G,G]\to H$ such that $f=g\circ\pi$ where $\pi:G\to G/[G,G]$ is the projection. This universal property gives you bijection between homomorphisms $G\to H$ and $G/[G,G]\to H$ when $H$ is abelian.

In your specific case, where $G=S_4$, $[G,G]=A_4$ and $H=\mathbb{Z}_4$, you can show that $S_4/A_4\cong\mathbb{Z}_2$, so there are as many homomorphisms $S_4\to\mathbb{Z}_4$ as $\mathbb{Z}_2\to\mathbb{Z}_4$. Clearly there are only two homomorphisms $\mathbb{Z}_2\to\mathbb{Z}_4$, since you either send $1$ to $0$ or $2$. This already proves the claim, but you can further check that these correspond, by the universal property, to the trivial homomorphism and the non-trivial one that you defined, respectively.

Final comment: these are all basic results in algebra, so, from what you wrote in your question, they probably summed up all of this to "$A_4$ must be in the kernel of your homomorphism $f$", which is equivalent to saying that there exists another homomorphism $g:S_4/A_4\to\mathbb{Z}_4$ such that $f=g\circ\pi$.

I hope this helps.

Paulo Mourão
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