Consider the homomorphism theorem:

Let $\phi:G \rightarrow\bar{G}$ be a surjective homomorphism, and let $N = ker(\phi)$. Let $\pi:G \rightarrow G/N$ be the quotient homomorphism. Then there exists unique group isomorphism $\bar \phi:G/N \rightarrow \bar{G}$ such that $\bar \phi \circ \pi = \phi$

I am trying to use this to find an isomorphism between $S_4/H$ and $S_3$ where $H=\{e, (12)(34), (13)(24), (14)(23)\}$.

I proved that $H$ is a normal subgroup, so we know that $\pi$ exists and it is very easy to compute. Now I'm trying to find a surjective homomorphism $\phi: S_4 \rightarrow S_3$ such that the kernel is H. Am I on the right track to finding this isomorphism and how do I find an $\phi$ with $H$ as the kernel?


There is a beautiful (and helpful) illustration of this $\phi$ in Garling's book on Galois Theory (CUP), in the introduction on groups: Label 4 vertices forming a scalene quadrilateral in the plane with 1,2,3,4. Draw all six straight lines through pairs (i,j) of them, labelling the lines by [ij]. Now label the three intersections of a pair of straight lines without a common point of 1,2,3,4 in the obvious way (i.e. line [12] intersects line [34] in point [12][34]). Now you can see that each permutation $\sigma$ of {1,2,3,4} (and according change of all labels) acts on the three points [12][34], [13][24], and [14][23]. Call the induced permutation $\phi (\sigma)$ and check that $\phi$ is a homomorphism with kernel H.


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