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If I want to write $(123)$ as product of transpositions, I get $(13)(12)$. For $(132)$ I get $(12)(13)$. For $(1234)$, I get $(14)(13)(12)$. Seems like I can write $(abcd)$ as $(ad)(ac)(ab)$.

Is this is a general trick that also works for larger $n$?

Edit: As usernull suggested, would this be a correct prove by induction?

Suppose that $(a_1 ... a_n)=(a_1 a_n)...(a_1 a_2)$. Then \begin{align} (a_1 ... a_n a_{n+1})=&(a_1 a_{n+1})(a_1 ... a_n) \\ \overset{\text{IH}}{=}& (a_1 a_{n+1})(a_1 a_n)...(a_1 a_2) \end{align}

Kasper
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3 Answers3

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$(1)$ Yes, as mixedmath points out, one "tried and true" method for writing a product of disjoint cycle as a product of transpositions is the strategy you employ:

For a one-cycle:
$$(a_1\,a_2\,a_3\,\cdots\,a_{n-1}\,a_n) = (a_1\,a_n)(a_1\,a_{n-1})\cdots\,(a_1\,a_3)(a_1\,a_2)$$

For more cycles, say e.g., a three-cycle, you just concatenate each cycle's product of transpositions: $$(abcd)(efgh)(ijkl) = \underbrace{(ad)(ac)(ab)}_{\large (abcd)}\,\underbrace{(eh)(eg)(ef)}_{\large (efgh)}\,\underbrace{(il)(ik)(ij)}_{\large(ijkl)}$$

But another fool-proof way to write cycles as products of transpositions is as follows:

$$(a_1\,a_2\,a_3\,\cdots\,a_{n-1}\,a_n) = (a_1\,a_2)(a_2\,a_3)\,(a_3\,a_4)\cdots(a_{n-2}\,a_{n-1})(a_{n-1}\,a_n)$$

And again, for more cycles, say e.g., a three-cycle, you just concatenate each cycle's product of transpositions: $$(abcd)(efgh)(ijkl) = \underbrace{(ab)(bc)(cd)}_{\large (abcd)}\,\underbrace{(ef)(fg)(gh)}_{\large (efgh)}\,\underbrace{(ij)(jk)(kl)}_{\large(ijkl)}$$

  • What this shows is that there is no unique way to write a permutation as a product of transpositions. What is true is that no matter which method you use, if it's a correct method, will yield the original cycle(s), when those transpositions are "multiplied out".

  • AND for every permutation, whatever the way chosen to express it as of the product of transpositions, the number of transpositions in the product will ALWAYS be even, or ALWAYS be odd, for any given permutation.

amWhy
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  • aah thanks! I like your other fool proof way to write cycles as products of permutations. Is it true that you could write this also the other way around? So: $(a_{n-1} a_n)...(a_1 a_2)$ the same as $(a_1 a_2)...(a_{n-1} a_n)$ – Kasper Mar 16 '13 at 16:04
  • No, it's not.: see what you get, for example, writing $(123) = (12)(34)$: when you multiply the transpositions, you get back $(123)$. But trying to write it as $(23)(12)$ gives you $(132)$ – amWhy Mar 16 '13 at 16:10
4

Yes, it always works.

$(abcdefg) = (ag)(af)(ae)(ad)(ac)(ab)$, and so on, will always work. One way to 'see it' is to justify that each letter goes to the next letter. Say we want to see what happens to $d$. Then in our product of transpositions, $d$ will be sent to $a$, and in the subsequent transposition $a$ is sent to $e$. Each letter besides $a$ appears exactly once, so we see that $d \to e$. This happens for the $n$th symbol in the exact same way.

davidlowryduda
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1

For any disjoint cycle $(a_1 a_2 ... a_n)$, you can write it as $(a_1 a_2)(a_1 a_3) \dots (a_1 a_n)$.

For a general permutation, represent it in disjoint cycle notation and then apply the above trick.

Edit: It appears you are evaluating your permutations with arguments on the right, in which case $(a_1 a_2 \dots a_n) = (a_1 a_n) \dots (a_1 a_3) (a_1 a_2)$ as you say.

muzzlator
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