Finding the homomorphisms $S_3 \to \Bbb Z / 6 \Bbb Z$

Question

I have to find explicitly (i.e. as operated on the element of the domain) the homomorphisms (of groups) from the symmetric group $S_3$ to $\Bbb Z / 6 \Bbb Z$.

Do I study the possible kernels of the homomorphisms?

Answer 1

Hint Since $\Bbb Z / 6 \Bbb Z$ is abelian, we must have for any homomorphism $\phi: S_3 \to \Bbb Z / 6 \Bbb Z$ that (for all $a, b \in S_3$) $$\phi(a b a^{-1} b^{-1}) = \phi (a) + \phi(b) - \phi(a) - \phi(b) = 0.$$ Thus, $\ker \phi$ contains the (normal) commutator subgroup $[S_3, S_3]$ of $S_3$, and hence $\phi$ descends to a homomorphism $S_3 / [S_3, S_3] \to \Bbb Z / 6 \Bbb Z$. (This applies to all group homomorphisms $G \to H$ with $H$ abelian, in fact.) Can you identify what $[S_3, S_3]$ is?