Klein Four-Group is Normal in A4
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Theorem
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
- $\begin{array}{c|cccc|cccc|cccc} \circ & e & t & u & v & a & b & c & d & p & q & r & s \\ \hline e & e & t & u & v & a & b & c & d & p & q & r & s \\ t & t & e & v & u & b & a & d & c & q & p & s & r \\ u & u & v & e & t & c & d & a & b & r & s & p & q \\ v & v & u & t & e & d & c & b & a & s & r & q & p \\ \hline a & a & c & d & b & p & r & s & q & e & u & v & t \\ b & b & d & c & a & q & s & r & p & t & v & u & e \\ c & c & a & b & d & r & p & q & s & u & e & t & v \\ d & d & b & a & c & s & q & p & r & v & t & e & u \\ \hline p & p & s & q & r & e & v & t & u & a & d & b & c \\ q & q & r & p & s & t & u & e & v & b & c & a & d \\ r & r & q & s & p & u & t & v & e & c & b & d & a \\ s & s & p & r & q & v & e & u & t & d & a & c & b \\ \end{array}$
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
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- $\index {A_4} V = 3$ follows from Lagrange's Theorem.
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.7$. Quotient groups: Example $127$