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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.

example

  • 1
    What is $Z_4$ ?2012-12-29
  • 0
    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 1.2012-12-29
  • 0
    @Amr I believe that means $V_4$, and it's a typo.2012-12-29
  • 0
    @Calvin Lin . OK this makes sense2012-12-29

3 Answers 3