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Let $n \ge 2$, $n \in \mathbb N$ and set $\sigma=(1,2,3,...,n)$ and $\tau=(1,2)$. Show that $\mathrm{Sym}(n)=\left<\sigma,\tau\right>$.

Approach: Induction

Proof:

Base case $n=2$

$\sigma=(1,2)$
$\tau=(1,2)$

$Sym(2)=\{Id_2,(1,2)\}$ $(1,2)=\tau$ and $Id_2=\tau\sigma$

so base case holds

Inductive step assume $Sym(k)=<\sigma,\tau>$ where $\sigma=(1,2,3,4,...,k)$ and $\tau=(1,2)$

Show $Sym(k+1)=<\lambda,\tau>$ where $\lambda={1,2,...,k+1}$ and $\tau=<1,2>$

To prove the inductive step, I was thinking we have to use the fact that every $x\in Sym(k+1)$ can be represented as the product of pairwise disjoint cycles. I think that we can express $\lambda$ in terms of $\sigma$

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TheMathNoob
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1 Answers1

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Your previous question was about transpositions generating $S_n$. So we may try an approach where we attempt to generate all transpositions from the given two permutations.

First, conjugate $(12)$ by $(12\cdots n)$ repeatedly to obtain $(23),(34),\cdots,(n-1\,n)$.

This makes sense once you understand this: $\sigma(a_1\,a_2\,\cdots\,a_k)\sigma^{-1}=(\sigma(a_1)\,\sigma(a_2)\,\cdots\,\sigma(a_k))$.

Second, see if you can conjugate $(i\,i+1)$ by $(j\,j+1)$s systematically to get any $(i\,k)$. For example, conjugating $(12)$ by $(23)$ gives $(13)$, then conjugating that by $(34)$ gives $(14)$.

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  • I am quite lost, so we are trying to generate every transposition from $Sym(n)$? – TheMathNoob Aug 08 '16 at 07:20
  • That's the approach I said I'm discussing in my answer, yes. – anon Aug 08 '16 at 07:29
  • and you want me to get somehow $(12)^k(12*****n)^l=(23),(34),....,(n-1,n)$ – TheMathNoob Aug 08 '16 at 07:31
  • No. Do you know what conjugate means? – anon Aug 08 '16 at 07:32
  • I guess it means somehow to use different combinations of the two permutations to get some desired transposition? – TheMathNoob Aug 08 '16 at 07:33
  • No. Conjugating $\tau$ by $\sigma$ means going from $\tau$ to $\sigma\tau\sigma^{-1}$. – anon Aug 08 '16 at 07:34
  • so what do you mean conjugate $(1,2)$ by (1,2,,,,,,n)) repeatedly to obtain (23),(34),...,(n-1,n). Can you give me one example?. Thanks for helping me. I have been doing these type of exercises the last for days. This is the last one. My back is killing me. – TheMathNoob Aug 08 '16 at 07:37
  • One example: conjugating $(12)$ by $(12\cdots n)$ gives $(23)$. Using the fact I stated in my answer about conjugating cycle structures, $\sigma(a_1\cdots a_k)\sigma^{-1}=(\sigma(a_1)\cdots \sigma(a_k))$, this is because $(12\cdots n)$ applied to $1$ is $2$ and applied to $2$ is $3$, hence $(12)$ becomes $(23)$. – anon Aug 08 '16 at 07:38
  • how do you get $\sigma ^{-1}$? – TheMathNoob Aug 08 '16 at 07:41
  • I don't get $\sigma^{-1}$ because we don't need to calculate that. Like I already said, $$\sigma(a_1\cdots a_k)\sigma^{-1}=(\sigma(a_1)\cdots\sigma(a_k)),$$ so to calculate $(1\cdots n)(12)(1\cdots n)^{-1}$ we apply $(1\cdots n)$ to the labels $1$ and $2$ within the notation $(12)$. If you really want to know $\sigma^{-1}$, it's $$(12\cdots n)^{-1}=(n\cdots 21),$$ from reversing the order of the labels. – anon Aug 08 '16 at 07:43
  • got it, how do you get other transpositions different than (2,3)? – TheMathNoob Aug 08 '16 at 07:51
  • I said "repeatedly" conjugate in my answer. So then conjugate $(23)$ by $(1\cdots n)$ to get $(34)$, etc. – anon Aug 08 '16 at 07:52
  • Oh, wouldn't that do the job? because there is a theorem that states every cycle can be represented as the product of simple transpositions. – TheMathNoob Aug 08 '16 at 07:54
  • If you have that theorem to use then sure. – anon Aug 08 '16 at 07:56
  • ok, now the only thing left is to understand that weird theorem . – TheMathNoob Aug 08 '16 at 07:59