Find all homomorphisms $S_3 \rightarrow \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_3$ where $S_3$ is a three-element permutation group
I want to use normal subgroup of $S_3$ to do this task so I find that these are:
$$H_1=\left\langle \text{id}\right\rangle = \left\{ \text{id}\right\}$$
$$H_2=\left\langle (123)\right\rangle = \left\{ (123),(132),id\right\}=\left\langle (132)\right\rangle$$
$$H_3=S_3$$
However I don't know what I can do the next.
Best Answer
The group $S_3$ is not abelian, so there is not any injective homomorphism $S_3\to \mathbb{Z}_2\oplus \mathbb{Z}_2\oplus\mathbb{Z}_3$. Thus, any such homomorphism has non-trivial kernel. As the kernel is a normal subgroup of $S_3$ then the kernel of such homomorphism must be equal to $H_2$ or $H_3$. If the kernel is $H_3$ then we have the trivial homomorphism. If the kernel is $H_2$, then as every transposition has order 2, then transpositions must be send to the sumand $\mathbb{Z}_2\oplus \mathbb{Z}_2$. Recall that $S_3$ is generated by a transposition $(12)$ and a $3$-cycle so there are exactly 3 homomorphisms:
Finally, using that $(23)=(132)(12)(123)$ and $(13)=(23)(12)(23)$ you can determine explictly all the above homomorphims.