presentation of the symmetric group via transpositions fixing one element Consider the symmetric group $S_n$. If we use the most popular set of generators $\sigma_1, \sigma_2,\cdots,\sigma_{n-1}$ with $sigma_i$ being the transposition $(i \, i+1)$, it is well known that they generate $S_n$ with relations 
$$ \sigma_i ^2 = 1, $$ $$ \sigma_i \sigma_j = \sigma_j \sigma_i \,\mbox{with}\, j \ne i \pm 1, $$ and finally $$ (\sigma_i \sigma_j )^3 =1.$$
What is the situation if we use the generators $(1\, n), (2\, n),\cdots, (n-1 \, n)$? What is the minimal set of relations that these elements will generate $S_n$?
 A: I'm not sure if the relations are minimal, but I believe this is such a presentation:
$$
    B = \left\langle
        \tau_2, \dots, \tau_n
    \ :\ \begin{matrix}
        \tau_i^2=1 \text{ for all $i$},\\
        (\tau_i\tau_j)^3 \text{ for all $i\neq j$},\\
        (\tau_i \tau_j\tau_i\tau_k)^2=1 \text{ for distinct $i,j,k$}\\
    \end{matrix}\right\rangle.
$$
Write
$$
    A = \left\langle
        \sigma_2,\dots,\sigma_n
    \ :\ \begin{matrix}
        \sigma_i^2=1\text{ for all $i$},\\
        (\sigma_i\sigma_{i+1})^3=1 \text{ for $i < n$},\\
        \sigma_i\sigma_j = \sigma_j \sigma_i \text{ for $|i-j|\geq 2$}
    \end{matrix}\right\rangle
    \cong S_n.
$$
You can check that the maps $f\colon A \to B$ and $g\colon B\to A$ defined by
$$
    f(\sigma_i) = \begin{cases}
        \tau_2 & i=2\\
        \tau_{i-1}\tau_i\tau_{i-1} & i>2
    \end{cases}
    \qquad\text{and}\qquad
    g(\tau_i)
    = \sigma_2 \sigma_3 \cdots \sigma_{i-1}\sigma_i\sigma_{i-1}\cdots \sigma_3 \sigma_2
$$
are well-defined by showing that they preserve relations, and show that they are inverses.
