# Showing that the alternating group $A_n$ is generated by a specific set of 3-cycles

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.

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.
• Thank you for your answer. Is there a step by step way to begin with the expression $(1ab)$ and transform it into $(12b)^{-1}(12a)(12b)$? – booa Feb 28 '20 at 17:53
• 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?) – booa Feb 28 '20 at 17:57
• 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 $(1,2,x)$. – Arturo Magidin Feb 28 '20 at 18:02