Transpositions of Adjacent Elements generate Symmetric Group

Theorem
Let $n \in \Z: n > 1$.

Let $S_n$ denote the symmetric group on $n$ letters.

Then the transpositions $a_k = \begin{pmatrix} k & k + 1 \end{pmatrix}$ for $1 \le k < n$ are a set of generators for $S_n$.

They satisfy the relations:

Proof
First, we show that each $\begin{pmatrix} i & j \end{pmatrix}$ where $i < j$ is in the subgroup $\gen {a_1, a_2, \ldots, a_{n - 1} }$.

From Cycle Decomposition of Conjugate, we can conjugate $a_i$ by $a_{i + 1}$ to give:
 * $\begin{pmatrix} i & i + 2 \end{pmatrix}$

Conjugating $a_i$ by the product $a_{j - 1} a_{j - 2} \ldots a_{i + 1}$ will give:
 * $\begin{pmatrix} i & j \end{pmatrix}$

Next, we note that:
 * from K-Cycle can be Factored into Transpositions, every cycle is a product of transpositions.

and:
 * from Existence and Uniqueness of Cycle Decomposition, every permutation is a product of cycles.

Thus, every permutation can be obtained from some product of the given transpositions.

Thus $\gen {a_1, a_2, \ldots, a_{n - 1} }$ is a generator of $S_n$.

From Transposition is Self-Inverse, we have:
 * $a_k^2 = e$

From K-Cycle can be Factored into Transpositions, $a_i a_{i + 1}$ is the $3$-cycle $\begin{pmatrix} i & i + 1 & i + 2 \end{pmatrix}$.

Thus:
 * $\paren {a_k a_{k + 1} }^3 = e$

If $\size {i - j} > 1$, then from Disjoint Permutations Commute, $a_i$ and $a_j$ are disjoint.

Thus:
 * $\paren {a_i a_j}^2 = a_i^2 a_j^2 = e$