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I understand any permutation $(a_1 \text{ } a_2 ... a_k)$ in $S_n$ where $k \leq n$ can be written as a product of $(a_1 \text{ } a_k)(a_1a_{k-1})...(a_2 \text{ } a_1)$. Similarly, $(a_1a_2...a_k)$ can be written as $(1 \text{ } a_1)(1 \text{ } a_k)...(1 \text{ } a_2)(1 \text{ } a_1)$ (if the permutation has $1$ in it, it can be rearranged so $1$ is the first integer appearing on the left and then the first and last transpositions can be removed) which shows any permutation can be written as the products of $(1 \text{ } 2), (1 \text{ } 3),...,(1 \text{ } n)$.

Is it possible to write a permutation as a product of $(1 \text{ } 2),(2 \text{ } 3),...,(n-1 \text{ } n)$? If the permutation consists of consecutive numbers in increasing order then $(a_1 \text{ } a_2)...(a_{k-1} \text{ } a_k)$ works fine. But I cannot generalize it for any sequence of integers. Is this actually possible?

Edit

It turns out it is possible. I am hoping to generalize this claim so for any permutation, so I can write it as a product of $(1 2), (2 3), ...,(n-1 \text{ } n)$, the way I have shown it works for $(1 2), (1 3),...,(1 n)$. For instance if $\sigma =(352)$ then $\sigma=(1 3)(1 2)(1 5)(13)$. How can the same cycle be written as a product of $(1 2),(2 3),...(n-1 \text{ } n)?$

Josh
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  • As a matter of fact it is. You should find it proven in any group theory book. – Davide Trono Apr 14 '21 at 21:52
  • Any permutation can be written as a product of transpositions $(a_i a_j)$. So it is necessary and sufficient to show that any transposition $(a_i a_j)$ can be written as a product of transpositions $(1 2), (2, 3), \ldots, (n-1 n)$. WLOG assume $a_i < a_j$. Then $(a_i a_j) = (a_i a_i+1) \cdots (a_j - 1 a_j)$ (if you compose left to right, and in the reverse order if you compose right to left). Thus every transposition is such a product, and thus every permutation is such a product. – davidlowryduda Apr 14 '21 at 21:53
  • Does this answer your question? Ways of expressing permutations as products of transpositions – Shaun Apr 14 '21 at 21:54
  • @Shaun The link you provide implies it is possible. I am hoping to get a generalize this observation so for any permutation I can come up with a product of transpositions the way I have done it for $(1 2)...(1 n)$. I will edit the question to make it more clear. – Josh Apr 14 '21 at 22:13
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    The magic words are "bubble sort". – Igor Rivin Apr 14 '21 at 23:03
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    Imagine you have a bookshelf with books in a series, but they are all mixed up. You want to rearrange them into the proper order, but because they are heavy, you can only do so by swapping the position of two adjacent books. Can you rearrange them any way you want? Yes: move the first book all the way to the left by repeatedly swapping it with its neighbors until it is first. Then move the second one to its position the same way, etc. This is the “bubble sort” that Igor Rivin mentions, but perhaps that visualization will help you understand it. – Arturo Magidin Apr 15 '21 at 01:48
  • @ArturoMagidin definitely helps a lot with building the intuition. I will try to use this idea for my question. Thanks! – Josh Apr 15 '21 at 03:22
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    Bubble sort is the natural response from anyone who knows a tiny bit of algorithms. However, you do not need to know that. All you need is simple understanding of induction. – user21820 Apr 15 '21 at 09:14

1 Answers1

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Solution 1:

It is obvious that we can write any permutation as a product of transpositions.
So, we need only to show that any transposition is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
Let $1\leq i<j\leq n$.
$(i\text{ }j)=(i\text{ }i+1)\cdots(j-2\text{ }j-1)(j-1\text{ }j)\cdots(i+1\text{ }i+2)(i\text{ }i+1)$.
For example, $(3\text{ }7)=(3\text{ }4)(4\text{ }5)(5\text{ }6)(6\text{ }7)(5\text{ }6)(4\text{ }5)(3\text{ }4)$.

Solution 2:

Let $\sigma$ be any permutation in $S_n$.
Let $a:=\sigma(1)$.
If $a=1$, let $\tau_1:=\sigma$.
If $a\neq 1$, then, $\tau_1:=(2\text{ }1)\cdots(a-1\text{ }a-2)(a\text{ }a-1)\sigma$.
Then $\tau_1(1)=1$.

Let $b:=\tau_1(2)\in\{2,\dots,n\}$.
If $b=2$, let $\tau_2:=\tau_1$.
If $b\neq 2$, let $\tau_2:=(3\text{ }2)\cdots(b-1\text{ }b-2)(b\text{ }b-1)\tau_1$.
Then $\tau_2(1)=1, \tau_2(2)=2$.

Let $c:=\tau_2(3)\in\{3,\dots,n\}$.
If $c=3$, let $\tau_3:=\tau_2$.
If $c\neq 3$, let $\tau_3:=(4\text{ }3)\cdots(c-1\text{ }c-2)(c\text{ }c-1)\tau_2$.
Then $\tau_3(1)=1, \tau_3(2)=2,\tau_3(3)=3$.

$\vdots$

Let $y:=\tau_{n-2}(n-1)\in\{n-1,n\}$.
If $y=n-1$, let $\tau_{n-1}:=\tau_{n-2}$.
If $y\neq n-1$, let $\tau_{n-1}:=(n\text{ }n-1)\tau_{n-2}$.
Then $\tau_{n-1}(1)=1,\tau_{n-1}(2)=2,\dots,\tau_{n-1}(n-2)=n-2,\tau_{n-1}(n-1)=n-1$.

Let $z:=\tau_{n-1}(n)\in\{n\}$.
Then, $z=n$.
So, $\tau_{n-1}$ is the identity.

Therefore, the inverse of $\sigma$ is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
Since the inverse of the inverse of $\sigma$ is also a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$ and $\sigma$ is equal to the inverse of the inverse of $\sigma$, $\sigma$ is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.

tchappy ha
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