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

Question

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.

Answer 1

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.