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:
\(\ds \map {N_{S_3} } {\set e}\) | \(=\) | \(\ds S_3\) | ||||||||||||
\(\ds \map {N_{S_3} } {\set {e, \tuple {123}, \tuple {132} } }\) | \(=\) | \(\ds S_3\) | ||||||||||||
\(\ds \map {N_{S_3} } {\set {e, \tuple {12} } }\) | \(=\) | \(\ds \set {e, \tuple {12} }\) | ||||||||||||
\(\ds \map {N_{S_3} } {\set {e, \tuple {13} } }\) | \(=\) | \(\ds \set {e, \tuple {13} }\) | ||||||||||||
\(\ds \map {N_{S_3} } {\set {e, \tuple {23} } }\) | \(=\) | \(\ds \set {e, \tuple {23} }\) | ||||||||||||
\(\ds \map {N_{S_3} } {S_3}\) | \(=\) | \(\ds S_3\) |
Proof
The subgroups of $S_3$ are as follows:
The subsets of $S_3$ which form subgroups of $S_3$ are:
\(\ds \) | \(\) | \(\ds S_3\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set e\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {123}, \tuple {132} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {12} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {13} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {23} }\) |
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 $6$