If a can write a permutation $\sigma$ as a product like $\Delta \alpha \beta$, where $\Delta$ is a product of transpositions (in fact, anything) and $\alpha$ and $\beta$ are two disjoint transpositions, so the symbols moved by $\alpha$ and $\beta$ belong to the support of different cycles in disjoint cycle decomposition of $\sigma$?

Is this true? If so, does somebody have some clue of where to find a proof for that?

Many thanks for any help...


  • $\begingroup$ What do you mean by the transpositions being in different orbits of $\sigma$? Under what action? $\endgroup$ Commented Oct 7, 2013 at 13:22
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    $\begingroup$ Hmm, now it has been edited to read "in $\sigma$" instead, which makes even less sense to me. $\endgroup$ Commented Oct 7, 2013 at 13:26
  • $\begingroup$ Sorry if I misused the concepts, I am a newbie on Permutation Groups. When I used "different orbits in $\sigma$", I wanted to mean "different cycles of disjoint cycle decomposition of $\sigma$". $\endgroup$
    – LuizG
    Commented Oct 7, 2013 at 13:56
  • $\begingroup$ I made some changes to the question, perhaps now it's clearer. $\endgroup$
    – LuizG
    Commented Oct 7, 2013 at 13:59

1 Answer 1


No they don't: $(12)(13)(24) = (2413)$.

  • $\begingroup$ Could you explain how you interpret the question (what is meant by orbit in $\sigma$)? $\endgroup$ Commented Oct 7, 2013 at 13:31
  • $\begingroup$ If you imagine the symmetric group on $n$ letters acting on an $n$-element set, then I literally imagine the orbits of this group action. In more standard, elementary terms, I am thinking of the cycles when a permutation is written in standard terms, i.e. as a product of disjoint cycles. Of course, the question doesn't make perfect sense in this context—because a transposition cannot really belong to a cycle—but its elements can, and this is how I took the question. I thought of it this way basically because I think it's the best interpretation, not because I think it's very sensible. $\endgroup$ Commented Oct 7, 2013 at 13:48
  • $\begingroup$ Ah, but now I realize that I betrayed my own interpretation by failing to choose disjoint permutations. Let me fix that. $\endgroup$ Commented Oct 7, 2013 at 13:49
  • $\begingroup$ @TobiasKildetoft OK, fixed. (and I forgot to do the @ thing, so there's your notification) $\endgroup$ Commented Oct 7, 2013 at 13:57
  • $\begingroup$ Thanks @user33433! I think my question is answered, due to your counter example. $\endgroup$
    – LuizG
    Commented Oct 7, 2013 at 14:09

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