Is there is any specific rule of writing any permutation into product of its transpositions? I know that permutation $$(123...n)= (1, n)(1, n-1)(1, n-2) \cdots(12)$$ but how can we reach to this result. Please give me some idea about.
Permutation in product of its transpositions
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abstract-algebra
group-theory
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0On the right, just show that $i \mapsto i + 1$ for $i = 1, \ldots, n - 1$ (and deal with $n$ too). What's the first transposition that touches $i$? Where does it send $i$? Does that number get sent anywhere else by transpositions to the left of it, and so on? – 2017-02-03
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1@pjs36 I think the question was not so much to prove the result but to explain how we thought of it. Probably the answer is by trial and error. – 2017-02-03
2 Answers
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Yes there is.Take any permutation.
every permutation can be written as product of disjoint cycles
Say one such cycle is ($a_1,a_2,...a_n$) then u can write this cycle as ($a_1,a_n$)($a_1,a_{n-1}$)....($a_1,a_2$)
Repeat for each cycle and you have product of transpositions.
P.S. This representation may not be unique.But surely this is one way to achieve.
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An approach might be the following. (I am writing composition of functions right-to-left, as you do, although particularly with permutations I prefer it the other way around.)
$\alpha = (1 2 \dots n)$ maps $1$ to $2$, so we might start writing $$ \alpha = \cdots (1 2), $$ where the dots has yet to be filled with transpositions.
Now the left-hand side, as written so far, correctly maps $1$ to $2$, but maps $2$ to $1$, whereas $\alpha$ maps $2$ to $3$. So we fix this by adding $(1 3)$: $$ \alpha = \cdots (1 3) (1 2), $$ and now the left-hand side, as written so far, correctly maps $1$ to $2$, and $2$ to $3$, etc.
But then, as noted in another answer, this is by far not the only way of writing $\alpha$ as a product transpositions. One might for instance start with $$ \alpha = \cdots (2 3), $$ etc.