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

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

Solution 1:

In case you do not know, 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 $2$.

Solution 2:

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 $3$?