TITLE: generators and relations of symmetric group
QUESTION [8 upvotes]: Let $S_{n}$ be symmetric group. Then it is given by generators $\tau_{i}$ where $i=1,2,\ldots,n-1$ and relations
$${\tau_{i}}^2$$
$$\tau_{i}\tau_{j}=\tau_{j}\tau_{i}\text{ for }i\neq j\pm1$$ 
$$\tau_{i}\tau_{i+1}\tau_{i}=\tau_{i+1}\tau_{i}\tau_{i+1}$$
I will be pleased if someone can present me detailed proof of this fact.

REPLY [5 votes]: We shall refer to the above relations as the Moore relations.
Set $\tau_i=(i\ i+1)\in S_n$ for $i=1,\dots,n-1$. It's easily seen that these transpositions satisfy the Moore relations. 

Theorem. Let $G$ be a group defined by generators $\tau_1, \tau_2, \ldots, \tau_{n-1}$ and the Moore relations. Then $G\simeq S_n$.

Proof. 

It's well known that $S_n$ is a homomorphic image of $G$. If we show that $|G|\leq |S_n|$, then we are done.

Induction on $n$. There is nothing to prove for $n=1$. Suppose $n>1$. 

By the induction hypothesis, the subgroup $H=\langle\tau_2,\dots,\tau_{n-1}\rangle$ of $G$ satisfies $\left|H\right| \leq \left|S_{n-1}\right|$ (since $H$ is a homomorphic image of the group with generators $\tau_2,\ldots,\tau_{n-1}$ and the Moore relations without the ones that involve $\tau_1$), and therefore it is enough to prove that $[G:H]\le n$. 

Let $H_1=H, H_2=H_1\tau_1,\dots, H_n=H_{n-1}\tau_{n-1}$. We have $H_{i+1}\tau_i=H_i$ for $i=1,\dots,n-1$. Furthermore, if $i\neq j,j+1$, then $H_i\tau_j=H_i$. In order to show this let's set $\sigma_i=\tau_1\cdots\tau_{i-1}$ and note that $H_i=H\sigma_i$. 
If $j\ge i+1$, then $\tau_j$ commutes with $\sigma_i$ and we get 
$$H_i\tau_j=H\sigma_i\tau_j=H\tau_j\sigma_i=H\sigma_i=H_i.$$ 
(We have used $H\tau_j=H$ which is true because $\tau_j\in H$). 
If $1\le j<i-1$, then $\sigma_i=\sigma_j\tau_j\tau_{j+1}\sigma$, where $\sigma$ commutes with $\tau_j$. Thus $$H_i\tau_j=H\sigma_j\tau_j\tau_{j+1}\sigma\tau_j=H\sigma_j(\tau_j\tau_{j+1}\tau_j)\sigma=H\sigma_j(\tau_{j+1}\tau_j\tau_{j+1})\sigma=H\tau_{j+1}\sigma_j\tau_{j}\tau_{j+1}\sigma=H\sigma_j\tau_j\tau_{j+1}\sigma=H\sigma_i=H_i.$$ 
(We have used $H\tau_{j+1}=H$ which is true because $\tau_{j+1}\in H$).
Since $G=\langle\tau_1,\dots,\tau_n\rangle$, the relations above show that any right coset in $G$ modulo $H$ coincides with one of the $H_i$, and this is enough to conclude that $[G:H]\le n$.
Remark. I know this proof since I was a student, but I don't remember the source.