Symmetric Group on 3 Letters/Normalizers

Normalizers of the Subgroups of the Symmetric Group on 3 Letters

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

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

The normalizers of each subgroup of $S_3$ are given by:

 $\displaystyle \map {N_{S_3} } {\set e}$ $=$ $\displaystyle S_3$ $\displaystyle \map {N_{S_3} } {\set {e, \tuple {123}, \tuple {132} } }$ $=$ $\displaystyle S_3$ $\displaystyle \map {N_{S_3} } {\set {e, \tuple {12} } }$ $=$ $\displaystyle \set {e, \tuple {12} }$ $\displaystyle \map {N_{S_3} } {\set {e, \tuple {13} } }$ $=$ $\displaystyle \set {e, \tuple {13} }$ $\displaystyle \map {N_{S_3} } {\set {e, \tuple {23} } }$ $=$ $\displaystyle \set {e, \tuple {23} }$ $\displaystyle \map {N_{S_3} } {S_3}$ $=$ $\displaystyle S_3$

Proof

The subgroups of $S_3$ are as follows:

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

 $\displaystyle$  $\displaystyle S_3$ $\displaystyle$  $\displaystyle \set e$ $\displaystyle$  $\displaystyle \set {e, \tuple {123}, \tuple {132} }$ $\displaystyle$  $\displaystyle \set {e, \tuple {12} }$ $\displaystyle$  $\displaystyle \set {e, \tuple {13} }$ $\displaystyle$  $\displaystyle \set {e, \tuple {23} }$
$\map {N_{S_3} } {\set e}$ is the largest subgroup $N$ of $S_3$ in which $H$ is normal in $N$.
$\map {N_{S_3} } {\set e} = S_3$
$\map {N_{S_3} } {S_3} = S_3$

The index of $\set {e, \tuple {123}, \tuple {132} }$ is $2$.

Hence from Subgroup of Index 2 is Normal, $\set {e, \tuple {123}, \tuple {132} }$ is normal in $S_3$.

It follows that:

$\map {N_{S_3} } {\set {e, \tuple {123}, \tuple {132} } } = S_3$

From Normal Subgroups of Symmetric Group on 3 Letters, none of $\set {e, \tuple {12} }$, $\set {e, \tuple {13} }$ and $\set {e, \tuple {23} }$ is normal in $S_3$.

There are no larger subgroup of $S_3$ containing any of them, so they are their own normalizers.

$\blacksquare$