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Let $A_n$ be the alternating group for some integer $n\geq3$. For every distinct elements $i,j,k$ of $[1,n]$, let $\tau_{i,k,j}$ be the 3-cycle $(ikj)$. Let $$C=\{\sigma\ |\ (\exists i)(\exists j)(\exists k)(\{i,j,k\}\subset[1,n]\land|\{i,j,k\}|=3\land\sigma=\tau_{i,k,j})\}$$ and $$\mathcal{W}(C)=\bigcup_{m\in\mathbb{N}}\big\{\sigma\ |\ (\exists\upsilon)(\upsilon\in(C\cup C^{-1})^{[1,m]}\land\sigma=\prod_{i=1}^m\upsilon_i)\big\}.$$

Then $A_n=\mathcal{W}(C)$. Now, let $$D=\bigcup_{i=3}^n\{\tau_{1,2,i}\}.$$ I would like to show that $A_n=\mathcal{W}(D)$. I suppose this has to be done by induction: the base case is easy. However, I am not sure how to proceed for arbitrary $n$. Let $\sigma\in A_n$: then there exist $m\in\mathbb{N}$ and $\upsilon\in(C\cup C^{-1})^{[1,m]}$ such that $$\sigma=\prod_{i=1}^m\upsilon_i.$$ I now need to show that, for every $i\in[1,m]$, $\upsilon_i\in D$. But I am not sure how to do this.

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Note that if $\sigma_1,\sigma_2\in\mathcal{W}(D)$, then $\sigma_1^{-1}$ and $\sigma_1\sigma_2\in\mathcal{W}(D)$.

It is enough to show that every $3$-cycle lies in $\mathcal{W}(D)$. If $2\lt a\lt b\leq n$, then $(1,a,b) = (1,2,b)^{-1}(1,2,a)(1,2,b)\in \mathcal{W}(D)$. This gives all three cycles that contain $1$.

Finally, if $1\lt a\lt b\lt c$, then $(a,b,c) = (1,c,a)(1,c,b)(1,a,c)$. So $\mathcal{W}(D)$ contains all three cycles. Since it contains all $3$-cycles, $\mathcal{W}(C)\subseteq\mathcal{W}(D)\subseteq \mathcal{W}(C)$, so you are done.

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  • $\begingroup$ Thank you for your answer. Is there a step by step way to begin with the expression $(1ab)$ and transform it into $(12b)^{-1}(12a)(12b)$? $\endgroup$ – booa Feb 28 '20 at 17:53
  • $\begingroup$ @booa: I don't understand what you are asking. Can you explain? $\endgroup$ – Arturo Magidin Feb 28 '20 at 17:55
  • $\begingroup$ I meant to ask: Is there a general method by which you were able to decompose $(1ab)$ into $(12b)^{-1}(12a)(12b)$, or did you deduce it through trial and error? (I guess this is like asking: how did you come up with this particular decomposition?) $\endgroup$ – booa Feb 28 '20 at 17:57
  • $\begingroup$ If $\tau$ sends $a$ to $b$, then $\sigma\tau\sigma^{-1}$ sends $\sigma(a)$ to $\sigma(b)$. So to transform $(1,2,a)$ to $(1,a,b)$, we can just find a $3$-cycle $\sigma$ that sends $2$ to $1$ (so that $\sigma^{-1}$ will send $1$ to $2$), $a$ to $a$, and $1$ to $b$ (so that $\sigma(1,2,a)\sigma^{-1} = (b,1,a)=(1,a,b)$; to make it a $3$ cycle, we just need to send $b$ to $2$, giving $\sigma=(2,1,b)=(1,b,2)$. Then I used $\sigma^{-1}$ so it would be clear that it is a $3$-cycle of the form $(1,2,x)$. $\endgroup$ – Arturo Magidin Feb 28 '20 at 18:02
  • $\begingroup$ Ah great! thank you for your help. $\endgroup$ – booa Feb 28 '20 at 18:04

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