# Symmetric Group on 3 Letters/Normal Subgroups

## Normal Subgroups of the Symmetric Group on 3 Letters

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

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

Consider the subgroups of $S_3$:

The subsets of $S_3$ which form subgroups of $S_3$ are:

 $\ds$  $\ds S_3$ $\ds$  $\ds \set e$ $\ds$  $\ds \set {e, \tuple {123}, \tuple {132} }$ $\ds$  $\ds \set {e, \tuple {12} }$ $\ds$  $\ds \set {e, \tuple {13} }$ $\ds$  $\ds \set {e, \tuple {23} }$

Of those, the normal subgroups in $S_3$ are:

$S_3, \set e, \set {e, \tuple {123}, \tuple {132} }$

### Cayley table of Quotient Group

Let $H$ denote the normal subgroup $\set {e, \tuple {123}, \tuple {132} }$.

Let $K$ denote the coset of $H$ in $S_3$.

The Cayley table of the quotient of $S_3$ by $H$ is given as:

$\begin {array} {c|cc} S_3 / H & H & K \\ \hline H & H & K \\ K & K & H \end {array}$

## Proof

$S_3$ itself is normal in $S_3$ by Group is Normal in Itself.

$\set e$ is normal in $S_3$ by Trivial Subgroup is Normal.

$\set {e, \tuple {12} }$:

 $\ds \tuple {123} \tuple {13} \tuple {123}^{-1}$ $=$ $\ds \tuple {123} \tuple {13} \tuple {132}$ $\ds$ $=$ $\ds \tuple {123} \tuple {12}$ $\ds$ $=$ $\ds \tuple {23}$ $\ds$ $\notin$ $\ds \set {e, \tuple {12} }$

Hence $\set {e, \tuple {12} }$ is not normal in $S_3$.

$\set {e, \tuple {23} }$:

 $\ds \tuple {123} \tuple {23} \tuple {123}^{-1}$ $=$ $\ds \tuple {123} \tuple {23} \tuple {132}$ $\ds$ $=$ $\ds \tuple {123} \tuple {13}$ $\ds$ $=$ $\ds \tuple {12}$ $\ds$ $\notin$ $\ds \set {e, \tuple {23} }$

Hence $\set {e, \tuple {23} }$ is not normal in $S_3$.

$\set {e, \tuple {13} }$:

 $\ds \tuple {123} \tuple {13} \tuple {123}^{-1}$ $=$ $\ds \tuple {123} \tuple {13} \tuple {132}$ $\ds$ $=$ $\ds \tuple {123} \tuple {12}$ $\ds$ $=$ $\ds \tuple {23}$ $\ds$ $\notin$ $\ds \set {e, \tuple {13} }$

Hence $\set {e, \tuple {13} }$ is not normal in $S_3$.

$\set {e, \tuple {123}, \tuple {132} }$:

We have that $\set {e, \tuple {123}, \tuple {132} }$ is the set of even permutations of $S_3$.

Any permutation of the form $\alpha \pi \alpha^{-1}$, for $\pi$ even, is also even.

Thus:

$\forall \alpha \in S_3: \alpha \pi \alpha^{-1} \in \set {e, \tuple {123}, \tuple {132} }$

Hence $\set {e, \tuple {123}, \tuple {132} }$ is normal in $S_3$.

$\blacksquare$