Klein Four-Group is Normal in A4

Theorem
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:

The subsets of $A_4$ which form subgroups of $A_4$ are as follows:

Consider the order $4$ subgroup $V$ of $A_4$, presented by Cayley table:


 * $\begin{array}{c|cccc}

\circ & e & t & u & v \\ \hline e & e & t & u & v \\ t & t & e & v & u \\ u & u & v & e & t \\ v & v & u & t & e \\ \end{array}$

Then $V$ is normal in $A_4$.

Its index is:
 * $\index {A_4} V = \dfrac {\order {A_4} } {\order V} = \dfrac {12} 4 = 3$

The (left) cosets of $V$ are:
 * $V$
 * $A := a V$
 * $P := p V$

and the Cayley table of the quotient group $A_4 / V$ is given by:


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

\circ & V & A & P \\ \hline V & V & A & P \\ A & A & P & V \\ P & P & V & A \\ \end{array}$

Note that while $A_4 / V$ is Abelian, $A_4$ is not.

Proof

 * $\index {A_4} V = 3$ follows from Lagrange's Theorem.