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: {{:Alternating Group on 4 Letters/Cayley Table}}  The normality status of the non-trivial proper subgroups of $A_4$ is as follows: Order $2$ subgroups: {{begin-eqn}} {{eqn | l = T       | o = :=       | r = \set {e, t}       | c = Not normal }} {{eqn | l = U       | o = :=       | r = \set {e, u}       | c = Not normal }} {{eqn | l = V       | o = :=       | r = \set {e, v}       | c = Not normal }} {{end-eqn}} Order $3$ subgroups: {{begin-eqn}} {{eqn | l = P       | o = :=       | r = \set {e, a, p}       | c = Not normal }} {{eqn | l = Q       | o = :=       | r = \set {e, c, q}       | c = Not normal }} {{eqn | l = R       | o = :=       | r = \set {e, d, r}       | c = Not normal }} {{eqn | l = S       | o = :=       | r = \set {e, b, s}       | c = Not normal }} {{end-eqn}} Order $4$ subgroup: {{begin-eqn}} {{eqn | l = K       | o = :=       | r = \set {e, t, u, v}       | c = Normal }} {{end-eqn}}  == Proof == Testing one of the left cosets of $T = \set {e, t}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = a T       | r = \set {a \circ e, a \circ t}       | c =  }} {{eqn | r = \set {a, c}       | c =  }} {{eqn | l = T a       | r = \set {e \circ a, t \circ a}       | c =  }} {{eqn | r = \set {a, b}       | c =  }} {{eqn | o = \ne       | r = a T       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $T$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $U = \set {e, u}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = b U       | r = \set {b \circ e, b \circ u}       | c =  }} {{eqn | r = \set {b, c}       | c =  }} {{eqn | l = U b       | r = \set {e \circ b, u \circ b}       | c =  }} {{eqn | r = \set {b, d}       | c =  }} {{eqn | o = \ne       | r = b U       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $U$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $V = \set {e, v}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = c V       | r = \set {c \circ e, c \circ v}       | c =  }} {{eqn | r = \set {c, d}       | c =  }} {{eqn | l = V c       | r = \set {e \circ c, v \circ c}       | c =  }} {{eqn | r = \set {c, b}       | c =  }} {{eqn | o = \ne       | r = c V       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $V$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $P = \set {e, a, p}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = t P       | r = \set {t \circ e, t \circ a, t \circ p}       | c =  }} {{eqn | r = \set {t, b, q}       | c =  }} {{eqn | l = P t       | r = \set {e \circ t, a \circ t, p \circ t}       | c =  }} {{eqn | r = \set {t, c, s}       | c =  }} {{eqn | o = \ne       | r = t P       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $P$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $Q = \set {e, c, q}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = t Q       | r = \set {t \circ e, t \circ c, t \circ q}       | c =  }} {{eqn | r = \set {t, d, p}       | c =  }} {{eqn | l = Q t       | r = \set {e \circ t, c \circ t, q \circ t}       | c =  }} {{eqn | r = \set {t, a, r}       | c =  }} {{eqn | o = \ne       | r = t Q       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $Q$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $R = \set {e, d, r}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = t R       | r = \set {t \circ e, t \circ d, t \circ r}       | c =  }} {{eqn | r = \set {t, c, s}       | c =  }} {{eqn | l = R t       | r = \set {e \circ t, d \circ t, r \circ t}       | c =  }} {{eqn | r = \set {t, b, q}       | c =  }} {{eqn | o = \ne       | r = t R       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $R$ is not normal in $A_4$. {{qed|lemma}} Testing one of the left cosets of $S = \set {e, b, s}$ against its corresponding right coset: {{begin-eqn}} {{eqn | l = t S       | r = \set {t \circ e, t \circ b, t \circ s}       | c =  }} {{eqn | r = \set {t, a, r}       | c =  }} {{eqn | l = S t       | r = \set {e \circ t, b \circ t, s \circ t}       | c =  }} {{eqn | r = \set {t, d, p}       | c =  }} {{eqn | o = \ne       | r = t S       | c =  }} {{end-eqn}} The left coset does not equal the right coset and so $S$ is not normal in $A_4$. {{qed|lemma}} The cosets of $K = \set {e, t, u, v}$ are as follows: {{begin-eqn}} {{eqn | l = a K       | r = \set {a \circ e, a \circ t, a \circ u, a \circ v}       | c =  }} {{eqn | r = \set {a, c, d, b}       | c =  }} {{eqn | l = K a       | r = \set {e \circ a, t \circ a, u \circ a, v \circ a}       | c =  }} {{eqn | r = \set {a, b, c, d}       | c =  }} {{eqn | r = a K       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = p K       | r = \set {p \circ e, p \circ t, p \circ u, p \circ v}       | c =  }} {{eqn | r = \set {p, s, q, r}       | c =  }} {{eqn | l = K p       | r = \set {e \circ p, t \circ p, u \circ p, v \circ p}       | c =  }} {{eqn | r = \set {p, q, r, s}       | c =  }} {{eqn | r = p K       | c =  }} {{end-eqn}} The left cosets equal the right cosets and so $K$ is normal in $A_4$. {{qed|lemma}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Stabilizer of Polynomial|next = Normality Relation is not Transitive/Proof 1}}: Chapter $6$: Cosets: Exercise $11$