How do I find the number of group homomorphisms from $S_3$ to $\mathbb{Z}/6\mathbb{Z}$?

Question

How do I find the number of group homomorphisms from the symmetric group $S_3$ to $\mathbb{Z}/6\mathbb{Z}$?

Answer 1

Hint: The group $\mathbb{Z}/6\mathbb{Z}$ is Abelian, and $S_3$ is non-Abelian. What does this tell us about the kernel of any homomorphism from $S_3$ to $\mathbb{Z}/6\mathbb{Z}$? Can we guarantee that certain elements inside $S_3$ must lie in the kernel? How many should there be? Now use this to count the total number of homomorphisms.

Added: $S_3$ consists of three elements of order $2$, two elements of order $3$, and the identity. The elements of order $2$ and $3$ do not commute. What possible places in $\mathbb{Z}/6\mathbb{Z}$ could I send an element of order $2$? What about the two elements of order $3$?

Answer 2

The kernel of a homomorphism must be a normal subgroup of the inverse image. Simply $S_3$ has 3 normal subgroups which are $\{e\}$, $A_3$, and $S_3$.

Let $\phi : S_3 \rightarrow \mathbb{Z}_6 $.

Then possible kernels are $\{e\}$, $A_3$, and $S_3$.

Firstly, try $\{e\}$. By First Isomorphism Theorem, $S_3/{e}$ which is $S_3$ itself, $S_3\simeq \phi(S_3)$. The order of $S_3$ is 6 and observe that $\mathbb{Z}_6$ has the same order. Thus, it yields $\phi(S_3)=\mathbb{Z}_6$. However, $S_3$ is not abelian although $\mathbb{Z}_6$ is. It is contradiction. Therefore, $\ker\phi$ cannot be $\{e\}$.

Secondly, let's check for $\ker\phi=S_3$. Then you can map every element of $S_3$ to the identity of $\mathbb{Z}_6$, that is, $\phi(s)=0$, $\forall s \in S_3$.

The last option is $\ker\phi = A_3$, therefore the order of the factor group $S_3/A_3$ is $2$. First isomorphism theorem gives us: $S_3/A_3 \simeq \phi(S_3)$, then $\phi(S_3)$ is $\{3,0\}. $

$\phi(s)= 0$ if $s \in A_3$.

Otherwise, $\phi(s)= 3$

As a conclusion, the answer is $2$.

Answer 3

Let $\phi: G\rightarrow H$ be a surjective homomorphism. Suppose that $H$ is abelian. Can you prove that the commutator subgroup of $G$ must be contained in $\text{ker}\phi$? (Recall: the commutator subgroup of $G$, denoted $G'$, is the subgroup generated by $\{g^{-1}h^{-1}gh : g,h \in G \}$.)

By the first isomorphism theorem, $G/\text{ker}\phi\cong H$. Write $\text{ker}\phi$ as $K$. Since $H$ is abelian, so is $G/K$, and thus $(gK)(hK)=(hK)g(K) \Leftrightarrow (gh)K=(hg)K \Leftrightarrow (g^{-1}h^{-1}gh)K= K$ we have $g^{-1}h^{-1}gh\in K$. Since this is true for all $g,h\in G$, we have that $G'\leqslant K$. (Of course, we can simply do these steps backwards to show that the converse is true, so in fact $G'\leqslant K$ if and only if $G/K$ is abelian.)

By a simple computation, $S_3'$ is the subgroup of rotations of order $3$. I claim that $\phi:S_3\rightarrow \mathbb{Z}_6$ cannot be surjective. Why?

Since $[S_3:S_3']=2$ and $S_3'\leqslant \text{ker}\phi$, $S_3/\text{ker}\phi$ has order at most $2$.

With this information, you should be able to deduce which subgroups of $\mathbb{Z}_6$ can be an image of a homomorphism from $S_3$. From this you can easily count the number of homomorphisms.