# Count the number group homomorphisms from $S_3$ to $Z/6Z$ ?

I have to count the number of group homomorphisms from $S_3$ to $Z/6Z$ ?

1. 1
2. 2
3. 3
4. 6

I aware of the formula for calculating group homomorphisms defined on cyclic groups..here $S_3$ is not cyclic. Please suggest how to proceed.

Thank you so much Egbert for the much detailed and patient reply.

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Note that a homomorphism from $S_3$ to $\mathbb{Z}_6$ is a homomorphism into an abelian group. Therefore, there is a bijection

$$\mathrm{hom}(S_3,\mathbb{Z}_6)\simeq\mathrm{hom}(S_3/[S_3,S_3],\mathbb{Z}_6),$$

where $[S_3,S_3]$ is the normal subgroup of $S_3$ generated by the elements of the form $aba^{-1}b^{-1}$. Below it is demonstrated that $[S_3,S_3]=A_3$, so it follows from the fact that $A_3$ has index 2, that $$\mathrm{hom}(S_3,\mathbb{Z}_6)\simeq\mathrm{hom}(\mathbb{Z}_2,\mathbb{Z}_6).$$ There are only two elements in $\mathbb{Z}_6$ whose orders divide $2$, so it follows that there are only two homomorphisms from $S_3$ to $\mathbb{Z}_6$.

As Dylan pointed out in the comments, the subgroup $[G,G]$ of a group $G$ generated by the elements of the form $aba^{-1}b^{-1}$ is called the commutator subgroup of $G$. Note that each commutator of $S_n$ is an even permutation, and therefore $[S_n,S_n]$ is a subgroup of $A_n$. On the other hand, all the 3-cycles are commutators: $$(abc)=(ab)(ac)(ab)(ac).$$ Since the subgroup $A_3$ is generated by the 3-cycles, it follows that $[S_3,S_3]=A_3$. It follows that $S_3/[S_3,S_3]=\mathbb{Z}_2$.

About the bijection: note that every homomorphism from an arbitrary group into an abelian group always sends the commutators to 0. In other words, the kernel of any such homomorphism always contains the commutator subgroup. It holds in more generality: If $G$ is a group and $A$ is an abelian group, then there is a bijection $$\mathrm{hom}(G,A)\simeq\mathrm{hom}(G/[G,G],A)$$ More precisely, this bijection is given (from right to left) by precomposition with the quotient map $G\to G/[G,G]$. Thus, the bijection says the following: for every homomorphism $f:G\to A$ there exists a unique homomorphism $\tilde{f}:G/[G,G]\to A$ with the property that $f=\tilde{f}\circ\pi_{[G,G]}$. This is the universal property of the map (functor) $G\mapsto G/[G,G]$.

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Neat! This will work more generally, since one knows $[S_n, S_n]$. –  Dylan Moreland Jun 2 '12 at 13:52
@preeti This is the commutator subgroup. See in particular the section on abelianization. –  Dylan Moreland Jun 2 '12 at 13:57
@Egbert: Thanks so much. Please see the working above. –  preeti Jun 2 '12 at 15:19
I'm glad it was useful. Your part of the solution doesn't contain mistakes, but you could have said things a bit clearer. For example, why not just saying that $S_3/A_3$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ instead of winding around that fact? –  Egbert Jun 2 '12 at 15:33

You are looking for ways of embedding quotients of $S_3$ in $\mathbf Z/6\mathbf Z$. So try writing down

1. The normal subgroups of $S_3$ and the corresponding quotients.
2. The number of ways to embed each quotient from (1) in $\mathbf Z/6\mathbf Z$. For this it should be enough to note that a non-abelian quotient will refuse to live inside of $\mathbf Z/6\mathbf Z$, and that you can apply the formula you mentioned if the quotient happens to be cyclic.

As an example, $A_3 = \{e, [123], [132]\}$ is a normal subgroup of $S_n$, and the corresponding quotient $S_3/A_3$ is the group of order $2$. Does this occur as a subgroup of $\mathbf Z/6\mathbf Z$?

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You meant $A_{3}$ is a normal subgroup of $S_{3}$ –  S.C. Sep 16 at 14:46