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 $A_4$.
$\Box$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $6$: Cosets: Exercise $11$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 83 \alpha$