# Alternating Group on 4 Letters/Normality of Subgroups

## Normality of Subgroups of the Alternating Group on $4$ Letters

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 normality status of the non-trivial proper subgroups of $A_4$ is as follows:

 $\ds T$ $:=$ $\ds \set {e, t}$ Not normal $\ds U$ $:=$ $\ds \set {e, u}$ Not normal $\ds V$ $:=$ $\ds \set {e, v}$ Not normal
 $\ds P$ $:=$ $\ds \set {e, a, p}$ Not normal $\ds Q$ $:=$ $\ds \set {e, c, q}$ Not normal $\ds R$ $:=$ $\ds \set {e, d, r}$ Not normal $\ds S$ $:=$ $\ds \set {e, b, s}$ Not normal
 $\ds K$ $:=$ $\ds \set {e, t, u, v}$ Normal

## Proof

Testing one of the left cosets of $T = \set {e, t}$ against its corresponding right coset:

 $\ds a T$ $=$ $\ds \set {a \circ e, a \circ t}$ $\ds$ $=$ $\ds \set {a, c}$ $\ds T a$ $=$ $\ds \set {e \circ a, t \circ a}$ $\ds$ $=$ $\ds \set {a, b}$ $\ds$ $\ne$ $\ds a T$

The left coset does not equal the right coset and so $T$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $U = \set {e, u}$ against its corresponding right coset:

 $\ds b U$ $=$ $\ds \set {b \circ e, b \circ u}$ $\ds$ $=$ $\ds \set {b, c}$ $\ds U b$ $=$ $\ds \set {e \circ b, u \circ b}$ $\ds$ $=$ $\ds \set {b, d}$ $\ds$ $\ne$ $\ds b U$

The left coset does not equal the right coset and so $U$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $V = \set {e, v}$ against its corresponding right coset:

 $\ds c V$ $=$ $\ds \set {c \circ e, c \circ v}$ $\ds$ $=$ $\ds \set {c, d}$ $\ds V c$ $=$ $\ds \set {e \circ c, v \circ c}$ $\ds$ $=$ $\ds \set {c, b}$ $\ds$ $\ne$ $\ds c V$

The left coset does not equal the right coset and so $V$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $P = \set {e, a, p}$ against its corresponding right coset:

 $\ds t P$ $=$ $\ds \set {t \circ e, t \circ a, t \circ p}$ $\ds$ $=$ $\ds \set {t, b, q}$ $\ds P t$ $=$ $\ds \set {e \circ t, a \circ t, p \circ t}$ $\ds$ $=$ $\ds \set {t, c, s}$ $\ds$ $\ne$ $\ds t P$

The left coset does not equal the right coset and so $P$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $Q = \set {e, c, q}$ against its corresponding right coset:

 $\ds t Q$ $=$ $\ds \set {t \circ e, t \circ c, t \circ q}$ $\ds$ $=$ $\ds \set {t, d, p}$ $\ds Q t$ $=$ $\ds \set {e \circ t, c \circ t, q \circ t}$ $\ds$ $=$ $\ds \set {t, a, r}$ $\ds$ $\ne$ $\ds t Q$

The left coset does not equal the right coset and so $Q$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $R = \set {e, d, r}$ against its corresponding right coset:

 $\ds t R$ $=$ $\ds \set {t \circ e, t \circ d, t \circ r}$ $\ds$ $=$ $\ds \set {t, c, s}$ $\ds R t$ $=$ $\ds \set {e \circ t, d \circ t, r \circ t}$ $\ds$ $=$ $\ds \set {t, b, q}$ $\ds$ $\ne$ $\ds t R$

The left coset does not equal the right coset and so $R$ is not normal in $A_4$.

$\Box$

Testing one of the left cosets of $S = \set {e, b, s}$ against its corresponding right coset:

 $\ds t S$ $=$ $\ds \set {t \circ e, t \circ b, t \circ s}$ $\ds$ $=$ $\ds \set {t, a, r}$ $\ds S t$ $=$ $\ds \set {e \circ t, b \circ t, s \circ t}$ $\ds$ $=$ $\ds \set {t, d, p}$ $\ds$ $\ne$ $\ds t S$

The left coset does not equal the right coset and so $S$ is not normal in $A_4$.

$\Box$

The cosets of $K = \set {e, t, u, v}$ are as follows:

 $\ds a K$ $=$ $\ds \set {a \circ e, a \circ t, a \circ u, a \circ v}$ $\ds$ $=$ $\ds \set {a, c, d, b}$ $\ds K a$ $=$ $\ds \set {e \circ a, t \circ a, u \circ a, v \circ a}$ $\ds$ $=$ $\ds \set {a, b, c, d}$ $\ds$ $=$ $\ds a K$

 $\ds p K$ $=$ $\ds \set {p \circ e, p \circ t, p \circ u, p \circ v}$ $\ds$ $=$ $\ds \set {p, s, q, r}$ $\ds K p$ $=$ $\ds \set {e \circ p, t \circ p, u \circ p, v \circ p}$ $\ds$ $=$ $\ds \set {p, q, r, s}$ $\ds$ $=$ $\ds p K$

The left cosets equal the right cosets and so $K$ is normal in $A_4$.

$\Box$