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

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:

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 $S_3$.