I am trying to show that $S_n = \langle \{(i,i+1) ~|~ i \in [1,n-1] \} \rangle = Q$.
I have that $S = \langle \{(i,j) ~|~ 1 \le i < j \le n-1\} \rangle = \langle T \rangle$ , so my goal is to show that $T \subseteq Q$. I discovered that $(i,i+2) = (, and so my hope is that this serves a base case. Through some examples I made the following conjecture:
$$(i,i+k) = \bigg(i+(k-1),i+k \bigg) \bigg(i+(k-2),i+(k-1) \bigg)...\bigg(i,i-1 \bigg)...\bigg(i+(k-2),i+(k-1) \bigg(i+(k-1),i+k \bigg)$$
Assuming that the formula holds for some natural number $k$,
$$\bigg(i,i+(k+1) \bigg) = \bigg(i,i+k \bigg) \bigg(i,i+(k+1),i+k \bigg)$$
This is where I got stuck. I am presently blanking: how can we write $\bigg(i,i+(k+1),i+k \bigg)$ as a product of "consecutive" transpositions? My other question is, is there a simpler way to solve this problem?
EDIT:
Okay. I have since taken a look at the duplicate and come to realize that first proving $S_n$ can be generated by $\{(1,2),(1,3),...,(1,n)\}$ makes things considerably simpler I did this exactly as Keith Conrad did it in his proof of theorem 2.2.
However, I think I might have proved theorem 2.3 (which is the problem I initially posed) slightly differently, but hopefully someone will correct me if I am wrong. I am not sure it is correct, so here goes nothing.
First, notice that $(1,3) = (2,3)(1,2)(2,3)$ and therefore $(1,3)$ is in $Q$. Now, suppose that $(1,i) \in Q$ for all integers up to some $i \in [1,n-1]$. Then $(1,i+1) = (i,i+1)(1,i)(i,i+1)$ Since $(i,i+1) \in Q$ and $(1,i)$ is in $Q$ by hypothesis, it follows that $(1,i+1) = (i,i+1)(1,i)(i,i+1) \in Q$, and therefore $S_n = \langle \{(1,2),...,(1,n)\} \rangle \subseteq Q$.
This seems a little simpler than the proof given by Keith Conrad, which can be found the aforementioned link.
