TITLE: $S_4/V_4$ isomorphic to $S_3$ - Understanding Attached Tables
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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)).