Symmetric Group on 4 Letters/Subgroups/Examples/Disjoint Transpositions

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Example of Subgroup of Symmetric Group on 4 Letters

Let $H$ be the subset of the Symmetric Group on $4$ Letters $S_4$ which consists of the $3$ products of disjoint transpositions of $S_4$, and the identity:

$V := \set {e, \tuple {1 2} \tuple {3 4}, \tuple {1 3} \tuple {2 4}, \tuple {1 4} \tuple {2 3} }$

Then $V$ forms a subgroup of $S_4$.


The Cayley table of $V$ can be presented as:

$\begin{array}{c|cccc}

\circ & e & \tuple {1 2} \tuple {3 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} \\ \hline 

                       e &                         e & \tuple {1 2} \tuple {3 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} \\

\tuple {1 2} \tuple {3 4} & \tuple {1 2} \tuple {3 4} & e & \tuple {1 4} \tuple {2 3} & \tuple {1 3} \tuple {2 4} \\ \tuple {1 3} \tuple {2 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} & e & \tuple {1 2} \tuple {3 4} \\ \tuple {1 4} \tuple {2 3} & \tuple {1 4} \tuple {2 3} & \tuple {1 3} \tuple {2 4} & \tuple {1 2} \tuple {3 4} & e \\ \end{array}$


This is the Klein $4$-group.

$V$ is normal in $S_4$.

The quotient group $S_4 / V$ is the Symmetric Group on $3$ Letters $S_3$.


The (left) cosets of $V$ are:

\(\ds E = e V\) \(=\) \(\ds \set {e, \tuple {1 2} \tuple {3 4}, \tuple {1 3} \tuple {2 4}, \tuple {1 4} \tuple {2 3} }\)
\(\ds A = \tuple {1 2 3} V\) \(=\) \(\ds \set {\tuple {1 2 3}, \tuple {1 3 4}, \tuple {2 4 3}, \tuple {1 4 2} }\)
\(\ds B = \tuple {1 3 2} V\) \(=\) \(\ds \set {\tuple {1 3 2}, \tuple {2 3 4}, \tuple {1 2 4}, \tuple {1 4 3} }\)
\(\ds C = \tuple {2 3} V\) \(=\) \(\ds \set {\tuple {2 3}, \tuple {1 3 4 2}, \tuple {1 2 4 3}, \tuple {1 4} }\)
\(\ds D = \tuple {1 3} V\) \(=\) \(\ds \set {\tuple {1 3}, \tuple {1 2 3 4}, \tuple {2 4}, \tuple {1 4 3 2} }\)
\(\ds F = \tuple {1 2} V\) \(=\) \(\ds \set {\tuple {1 2}, \tuple {3 4}, \tuple {1 3 2 4}, \tuple {1 4 2 3} }\)

and its resulting Cayley table is:

$\begin{array}{c|ccc|ccc}

\circ & E & A & B & C & D & F \\ \hline E & E & A & B & C & D & F \\ A & A & B & E & F & C & D \\ B & B & E & A & D & F & C \\ \hline C & C & D & F & E & A & B \\ D & D & F & C & B & E & A \\ F & F & C & D & A & B & E \\ \end{array}$


Proof

We have that $V$ contains all the elements of $S_4$ of the same cycle type.

From Cycle Decomposition of Conjugate, the conjugate of a permutation is another permutation of the same cycle type.

Hence the conjugate of an element of $V$ is an element of $V$.

That is, $V$ is normal in $S_4$.

The quotient group $S_4 / V$ is of order $\dfrac {24} 4 = 6$, so must be either $C_6$ or $S_3$.

From the Cayley table above it is apparent that it is $S_3$.

$\blacksquare$


Sources

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