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If I understood correctly, any permutation can be written as a sequence $s$ ("product") of distinct transpositions (2-element swaps, each swap occurs at most once in the sequence (I consider 2 swaps equal if they swap the elements at the same positions, no matter which elements occupy the position at that time in the sequence)). For example, consider the permutation $(3,4,2,1)$ which can be obtained by swapping positions $(1,4)$, $(2,3)$, and then $(1,2)$ (at that point positions 1 and 2 are occupied by elements 4 and 3).

Now, I'm very uncertain about the degree to which I can reorder the transpositions without changing the resulting permutation. For example $(3,4,2,1)$ can also be obtained by swapping $(2,3), (1,4),(1,2)$, but not by $(1,2),(1,4),(2,3)$.

In particular, I want to assume without loss of generality that, for any given permutation, there is a sequence $s$ of transpositions such that the following transposition $(i,j)$ is last in $s$: here, $j$ is the maximum element and $i$ is the maximum over all elements occuring with $j$ in a transposition in $s$.

Can I assume this without loss of generality and what should I cite to support this? Can I assume other properties about the sequence (for example, could all transpositions be of the form $(1,j)$ while maintaining the property that none of them is repeated)?

igel
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1 Answers1

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Any permutation breaks down into disjoint cycles. For example $(3,1,2,5,4)$ is the composition of cycles $(123)(45)$. The cycles can be performed in any order and we will get the same permutation.

Further each cycle can be broken down into a product of transpositions of the form $(a_ka_{k-1})(a_{k-1}a_{k-2})\cdots (a_2a_1)$, with the $a_i$ distinct. For example: $$(12345)=(54)(43)(32)(21).$$ Given a cycle, we can label any of its elements $a_1$, so we can get the final transposition $(a_2a_1)$ in the factorisation to be any transposition of consecutive elements in the cycle.

Thus given a permutation, you can express it as a product of transpositions, with the final transposition being any transposition of consecutive elements in a cycle that you want.

Note that this factorisation into transpositions does not use any transposition more than once.

We have $$(ij)=(1i)(1j)(1i),$$ so any permutation can be expressed as a product of $(1i)$. However it may that you need to use the same transposition more than once. For example the permutation $(1,3,2)$ is not equal to $(12)(13)$ or $(13)(12)$.

tkf
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  • "Note we can cycle the order of these transpositions around, and still get the same cycle." - this is interesting. But (12)(14)(23) = (4312) and (14)(23)(12) = (3421) but (12)(14)(23) is just a cyclic shift of (14)(23)(12), no? Maybe I misunderstood something... – igel Dec 20 '21 at 21:05
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    The key word was "these", but I should edit to make this clearer. I was referring specifically to products of the form $(a_ka_{k-1})(a_{k-1}a_{k-2})\cdots (a_2a_1)$ with the $a_i$ distinct. Now edited. Does it make sense now? – tkf Dec 20 '21 at 21:16
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    Cycles commute iff they are disjoint. – Javi Dec 20 '21 at 22:33
  • I think I got it now, thanks! One last thing, is this cyclic shifting equivalence of this particular representation folklore in algebra? That is, do I just cite any introductional monograph for this or is there a particular paper that proves it? – igel Dec 20 '21 at 22:53
  • I think you just say it holds by symmetry: The cycle $(a_1a_2\cdots a_k)$ is invariant under cycling round the $a_i$, so you can cycle round the $a_i$ in the factorisation $(a_ka_{k-1})(a_{k-1}a_{k-2})\cdots (a_2a_1)$ which is the same as cycling round the transpositions. – tkf Dec 21 '21 at 00:21
  • IMHO we should do the following correction: 1. By cycle definition https://en.wikipedia.org/wiki/Cyclic_permutation#Definitio, (123)(45) should be (132)(45). 2. Maybe you want to say "you can put any cycle you like last" which corresponds to "give you the last cycle" – An5Drama Feb 16 '24 at 03:37
  • Here are some clarification: 1. "Note (in the case of products as above" may mean $(a_ka_{k-1})(a_{k-1}a_{k-2})\cdots (a_2a_1)$ can be written as $(a_2a_1)(a_3 a_2)\cdots (a_{k-1}a_{k-2})(a_ka_{k-1})$. This is shown in https://en.wikipedia.org/wiki/Cyclic_permutation#Properties and https://math.stackexchange.com/a/3420633/1059606 (Notice we should execute the right factor first as wikipedia says). 2. "does not use any transposition more than once" is due to permutation is decomposed into one or more disjoint cycles. – An5Drama Feb 16 '24 at 03:37