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Consider this equality in the group $S_n$ :


I can prove it for myself, but I get the impression that I should be able to see this very quickly. How could I get more intuition on this ?

$$ a \overset{(abc)}{\mapsto} b \overset{(cad)}{\mapsto} b \\ b \overset{(abc)}{\mapsto} c \overset{(cad)}{\mapsto} a \\ c \overset{(abc)}{\mapsto} a \overset{(cad)}{\mapsto} d \\ d \overset{(abc)}{\mapsto} d \overset{(cad)}{\mapsto} c \\$$

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My adivinatory skills tell me that you're talking of...cycles in the permutation group? ... – DonAntonio Mar 22 '13 at 16:12
A seemingly small edit came up for approval, in which someone changed "in $S_n$" to "in a group $S_n$". I clicked on "improve" and changed it to "in the group $S_n$", and didn't notice that the title had also been altered to one that doesn't make sense. I suspect the comments from DonAntonio and Hurkyl are responding to the nonsensical title. So I changed the title back to what it was originally. – Michael Hardy Mar 22 '13 at 16:24
@DonAntonio : Please notice my comment and consider deleting yours. – Michael Hardy Mar 22 '13 at 16:25
@Hurkyl : Please notice my comment and consider deleting yours. – Michael Hardy Mar 22 '13 at 16:25
Sorry but no, can't do Michael: I require the OP to at least acknowledge the editing is correct, and then I will be happy to delete my comment, as I many times (when I remember!) do... – DonAntonio Mar 22 '13 at 16:27
up vote 7 down vote accepted

Instead of tracing through where it maps elements, you can also derive this from basic cycle manipulations: $$(cad)(abc) = (dca)(cab) = (dc)(ca)(ca)(ab) = (dc)(ab)$$

The two rules I'm using are

  1. $(a_1\ldots a_n) = (a_2\ldots a_na_1)$
  2. $(a_1\ldots a_n) = (a_1a_2)(a_2a_3)\cdots(a_{n-1}a_n)$
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thnx !! ${}{}{}{}{}{}$ – Kasper Mar 22 '13 at 16:27

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