Honestly I don't know how to come up with a way to explicitly write a surjective homomorphism $\phi: S_4 \to S_3$
What I do know (which is what I did):
- Showed that $S_4/H \cong S_3$, where $H$ is a normal subgroup of $S_4$.
- Since $|S_4/H| = \frac{|S_4|}{|H|} = |S_3| \implies \frac{24}{|H|} = 6 \implies |H| = 4$.
- Since a normal subgroup can be written as the union of (some) of the conjugacy classes of (in this case) $S_4$, I write them down and found out (after proving that $H$ is indeed a subgroup) that $H = \{e, (12)(34), (13)(24), (14)(23) \}$, which is the Klein-$4$ group (I didn't know about Klein groups before this problem (: ).
- Then by the theorem of homomorphisms $\exists$ surjective homomorphism $\phi: S_4 \to S_3$, such that $\mathrm{Ker}(\phi) = H$.
And here I'm stuck. By googling I found material (here and here), where showed that the (surjective) homomorphism exists (if I didn't misunderstood what it's written there) by conjugating the elements of $H$ by those of $S_3$. But
On what basis they decided that conjugation is a good idea to show that the homomorphism exists?
From what it's written there (if possible) how can one write down and explicit function $\phi$ or it's not necessary?
