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 A 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 A K-Cycle can be Factored into Transpositions. Thus $$\left({a_k a_{k+1}}\right)^3 = e$$.


 * If $$\left|{i - j}\right| > 1$$, then $$a_i$$ and $$a_j$$ are disjoint from Disjoint Permutations Commute. Thus $$\left({a_i a_j}\right)^2 = a_i^2 a_j^2 = e$$.