generators and relations of symmetric group 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. 
 A: 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. 
A: If I understand the question, it is to show that the symmetric group $S_n$ is exactly the universal group on those generators with those relations. That is, that it is the quotient of the free group on $n$ generators by the smallest normal subgroup containing the relators.
The potential issue would be that $S_n$ might be a proper quotient of the actual universal group with those relations.
One device to understand $S_n$ is as a Coxeter group: the actual universal group $W$ with generators $s_j$ and relations as in the question is by definition a Coxeter group, and the _reflection_representation_ $W\rightarrow O(V)$ to the orthogonal group of the corresponding invariant quadratic form (from the Coxeter data) is proven injective by consideration of the length function on the group. It is not entirely trivial to prove this (see the Corollary on p. 7 of http://www.math.umn.edu/~garrett/m/buildings/book.pdf, for example).
To see that $S_n$ is isomorphic to $W$, rather than to a proper quotient, we need a natural representation of $S_n$ on an $(n-1)$-dimensional vector space $V$ preserving a quadratic form isomorphic to the Coxeter form. Indeed, and unsurprisingly, the restriction of the Killing form to the diagonal matrices in the Lie algebra $sl(n)$ of $SL(n)$ is preserved by the conjugation (Adjoint) action of $n$-by-$n$ permutation matrices, and is a scalar multiple of the Coxeter quadratic form.
This may seem circuitous or long-winded, but it is fairly conceptual, and applies to concrete realizations of the (spherical) Weyl groups of the other classical groups, as well.
