Is any transposition a product of simple transpositions? If yes, how can you prove this?
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Every element of $S_n$ can be written as a product of simple transpositions. This fact can be proven by induction on $n$. As a first step, prove that $(1,m)$ is always a product of simple transpositions by induction of $m$. The case $m = 2$ is trivial, so we assume that $(1,m-1)$ is a product of simple transpositions. But $(1,m) = (m-1,m)(1,m-1)(m-1,m)$, so $(1,m)$ is a product of simple transpositions. Now we show that all transpositions in $S_n$ are products of transpositions of the form $(1,m)$. The case $n = 2$ is trivial, so we assume that any transposition of $S_{n-1}$ is such a product. Given the transposition $t$, we must have some $m$, possibly $1$ which exchanges position with 1. If we perform the transposition $(1,m)$, we only have $n-1$ positions left to permute, which can be done using simple transpositions by the inductive hypothesis. This completes the proof. |
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There is an algorithm that takes a permutation and writes it as a product of simple transpositions. It's called bubble sort. What you do is : if $f(i) > f(i+1)$, write $f$ = $f' \circ (i,i+1)$. Repeat on $f'$ until you get the identity and are left with a product of simple transposition. |
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Yes. One way to see this is to notice what happens when you conjugate transpositions by transpositions. For example, $(5,6)(3,6)(5,6)=(3,5)$ shows that you can get one step closer to a simple transposition, from $(3,6)$ to $(3,5)$, by conjugating by $(5,6)$. Then $(4,5)(3,5)(4,5)=(3,4)$, so we've gotten to a simple transposition after $2$ conjugations. To write $(3,6)$ as a product of simple transpositions, you can just unwind this: $(3,6)=(5,6)(4,5)(3,4)(4,5)(5,6)$. |
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