Group homomorphism between $S_n$ and $\mathbb{Z} / 3 \mathbb{Z}$

Question

So the question is:

What are the group homomorphism between $S_n$ and $\mathbb{Z} / 3 \mathbb{Z}$

Now my instinct would be to consider a homomorphism $f$ between $S_n$ and $\mathbb{Z}$ such that $\text{K} (f) = \mathbb{3Z}$, then I would be able to state that the exists a unique bijective $f'$ homomorphism between $\mathbb{Z}/3\mathbb{Z}$, and then I would need to construct the reciprocal function(which does exist as $f'$ is bijective). Now the issue is that I don't know what kind of $f$ I can construct. Can you help me on this?

And furthermore, is there any other way to proceed to get a different group homomorphism?

Answer 1

The group $S_n $ is generated by the transpositions $\sigma_{i,j}$, which are permutations that just swap two elements $i $ and $j $, leaving everything else unchanged. Any homomorphism $f $ of $S_n $ into $\mathbb Z/3\mathbb Z $ must map those transpositions into identity, because they are of order 2, and the only element in $\mathbb Z/3\mathbb Z $ that, squared, gives the identity is the identity itself. Therefore, as all the generators of $S_n $ must be mapped to identity, the homomorphism must map all the other elements to identity. Thus, every homomorphism of $S_n $ into $\mathbb Z/3\mathbb Z $ is trivial.

Answer 2

Hint: If $f: S_n \to \mathbb{Z} / 3 \mathbb{Z}$ is a homomorphism, then $\ker f$ is a normal subgroup of $S_n$. The normal subgroups of $S_n$ are well known. Moreover, the image of $f$ is a subgroup of $\mathbb{Z} / 3 \mathbb{Z}$ and so is either trivial or the whole group. Does $S_n$ have a normal subgroups of index $3$?

Answer 3

If $f$ is a homomorphism from $S_n$ to $\mathbb{Z}/3\mathbb{Z}$, the image is abelian. Hence the commutator subgroup $[S_n,S_n]=A_n \subseteq ker(f)$. But $|S_n:A_n|=2$, so $|S_n:ker(f)|$ must divide $2$ ... can you finish?

Answer 4

We have that $\ker \phi \unlhd S_n$. Let $\phi$ be the non-trivial homomorphism. Then we have by First Isomorphism Theorem that $S_n/\ker \phi \cong \phi[S_n] \cong \mathbb{Z_3}$. Now how many normal subgroup od index $3$ does $S_n$ have?