# Writing a permutation as products of transpositions

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...

Luiz

• What do you mean by the transpositions being in different orbits of $\sigma$? Under what action? – Tobias Kildetoft Oct 7 '13 at 13:22
• Hmm, now it has been edited to read "in $\sigma$" instead, which makes even less sense to me. – Tobias Kildetoft Oct 7 '13 at 13:26
• 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 $\sigma$". – LuizG Oct 7 '13 at 13:56
• I made some changes to the question, perhaps now it's clearer. – LuizG Oct 7 '13 at 13:59

## 1 Answer

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

• Could you explain how you interpret the question (what is meant by orbit in $\sigma$)? – Tobias Kildetoft Oct 7 '13 at 13:31
• 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. – Slade Oct 7 '13 at 13:48
• Ah, but now I realize that I betrayed my own interpretation by failing to choose disjoint permutations. Let me fix that. – Slade Oct 7 '13 at 13:49
• @TobiasKildetoft OK, fixed. (and I forgot to do the @ thing, so there's your notification) – Slade Oct 7 '13 at 13:57
• Thanks @user33433! I think my question is answered, due to your counter example. – LuizG Oct 7 '13 at 14:09