Find all homomorphisms $S_3 \rightarrow \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_3$

group-homomorphismgroup-theory

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:

  1. The one sending $(12)$ to $(1,0,0)$ and $H_2$ to $(0,0,0)$;
  2. The homomorphism sending $(12)$ to $(0,1,0)$ and $H_2$ to $(0,0,0)$; and
  3. The trivial homomorphism.

Finally, using that $(23)=(132)(12)(123)$ and $(13)=(23)(12)(23)$ you can determine explictly all the above homomorphims.

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