# 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:

\(\displaystyle \) | \(\) | \(\displaystyle S_3\) | |||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle \set e\) | |||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle \set {e, \tuple {123}, \tuple {132} }\) | |||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle \set {e, \tuple {12} }\) | |||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle \set {e, \tuple {13} }\) | |||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle \set {e, \tuple {23} }\) |

By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:

From Trivial Subgroup is Normal:

- $\map {N_{S_3} } {\set e} = S_3$

From Group is Normal in Itself:

- $\map {N_{S_3} } {S_3} = S_3$

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.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.11$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $6$