# Symmetric Group on 3 Letters/Generators

## Generators of the Symmetric Group on 3 Letters

Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$

Let:

 $\ds G_1$ $=$ $\ds \set {\tuple {123}, \tuple {12} }$ $\ds G_2$ $=$ $\ds \set {\tuple {13}, \tuple {23} }$

Then:

 $\ds S_3$ $=$ $\ds \gen {G_1}$ $\ds$ $=$ $\ds \gen {G_2}$

where $\gen G$ denotes the group generated by a subset $G$ of $S_3$.

## Proof

For $G_1$:

 $\ds e$ $=$ $\ds \tuple {12} \tuple {12}$ $\ds \tuple {123}$ $=$ $\ds \tuple {123}$ $\ds \tuple {132}$ $=$ $\ds \tuple {123} \tuple {123}$ $\ds \tuple {12}$ $=$ $\ds \tuple {12}$ $\ds \tuple {23}$ $=$ $\ds \tuple {123} \tuple {12}$ $\ds \tuple {13}$ $=$ $\ds \tuple {12} \tuple {123}$

For $G_2$:

 $\ds e$ $=$ $\ds \tuple {23} \tuple {23}$ $\ds \tuple {123}$ $=$ $\ds \tuple {13} \tuple {23}$ $\ds \tuple {132}$ $=$ $\ds \tuple {23} \tuple {13}$ $\ds \tuple {12}$ $=$ $\ds \tuple {13} \tuple {23} \tuple {13}$ $\ds \tuple {23}$ $=$ $\ds \tuple {23}$ $\ds \tuple {13}$ $=$ $\ds \tuple {13}$

$\blacksquare$