S4/V4 isomorphic to S3 - Understanding Attached Tables

I think I see $S_4/V_4 \cong S_3$ from the first table beneath marked in the green. I just ignore $V_4$ and think of it as mapped away by the bijection $f^{-1}$ where $f(s) = s V_4 \iff f^{-1}(\sigma V_4) = s \in S_3$ But why do they compute only $\{S_3\}V_4$? By definition, $S_4/V_4 = \{sV_4 : s \in S_4\}$. Where are the rest of the elements in $S_4/V_4$ like $(2, 1, 3, 4)V_4, (2, 1, 4, 3)V_4, (2, 3, 1, 4)V_4$ etc...?

I don't see "The rows are the cosets of $V_4$ in $S_4$." Can someone show me this please? For instance, the third row of the table marked in the blue consists of $(1, 4, 3, 2), (1, 3, 2, 4) \notin V_4$.

I can't see $S_4/V_4 \cong S_3$ from the second table. Can someone explain it please? Thank you.

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What is $Z_4$ ? –  Amr Dec 29 '12 at 23:37
Why do you say that $(1, 4, 3, 2) \not \in S_4$? It is the permutation on 4 elements that sends 1 to 4, 4 to 3, 3 to 2 and 2 to 1. –  Calvin Lin Dec 29 '12 at 23:43
@Amr I believe that means $V_4$, and it's a typo. –  Calvin Lin Dec 29 '12 at 23:43
@Calvin Lin . OK this makes sense –  Amr Dec 29 '12 at 23:46

I'm guessing that by $\,Z_4\,$ you actually mean the Klein viergrupp

$$C_2\times C_2\cong \{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$

since the above one is the only normal subgroup of order $\,4\,$ of $\,S_4\,$ , but then, since the cyclic group of order $\,6\,$ is obviously abelian ,we get

$$S_4/Z_4\cong C_6\Longleftrightarrow S_4^{'}\leq Z_4$$

and this is false since $\,(123)\notin Z_4\,$ , but $\,(123)\in A_4=S_4^{'}\,$.

Thus, as the only other group of order $\,6\,$, up to isomorphism, is $\,S_3\,$ ,we get that $\,S_4/Z_4\cong S_3\,$

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Thank you for your answer, sir. You wrote that since the cyclic group of order six is Abelian, the factor of $S_4$ by the Klein viergrupp is isomorphic to the said cyclic group exactly when the alternative group $S_{4}^{'}$ is a subgroup of the Klein group. Why is that? Are you referring to a more general statement? –  Student Jun 21 '14 at 17:03

A coset of $V_4$ is a set $gV_4=\{gv:v\in V_4\}$. So in fact any coset of $V_4$ formed by a $g\in G$ which is not in $V_4$ will contain no elements of $V_4$. (This is easy to prove; try it.)

DonAntonio has shown you a somewhat advanced way to see how $S_3\cong S_4/V_4$, but considering your questions for us, I think that the best way to see it, at this point, would be to write out the actual isomorphism $\mu:S_4/V_4\rightarrow S_3$. Use the cosets they've given you. Ensure you know how coset multiplication works - $(gV_4)(hV_4)=(gh)V_4$. You might have to go back and review some definitions.

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Where are the rest of the elements of $S_4/V_4$? Try computing one of them. You'll see that it's equal to one of the cosets already listed in the green section. (Remember that the order of elements in a set doesn't matter, and also that re-cycling the elements in a cycle also doesn't change it: (1,2,4,3) = (2,4,3,1) = (4,3,1,2) = (3,1,2,4)).

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