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

They satisfy the relations:

Poof
First, we show that each $\begin{bmatrix} i & j \end{bmatrix}$ where $i < j$ is in the subgroup $\left \langle {a_1, a_2, \ldots, a_{n-1}}\right \rangle$.

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

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

Next, we note that from K-Cycle can be Factored into Transpositions, every cycle is a product of transpositions, and from Cycle Decomposition, every permutation is a product of cycles.

Thus, every permutation can be obtained from some product of the given transpositions, thus $\left \langle {a_1, a_2, \ldots, a_{n-1}}\right \rangle$ is a generator of $S_n$.

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

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

Thus $\left({a_k a_{k+1}}\right)^3 = e$.

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

Thus $\left({a_i a_j}\right)^2 = a_i^2 a_j^2 = e$.