- Show that \(S_4\) has two normal subgroups: \(A_4, \ZZ_2^2\).
Solution
The proper nontrivial normal subgroups of \(S_4\) are \[ V_4=\{e,(12)(34),(13)(24),(14)(23)\} \] and \(A_4\).
Both are normal: \(A_4\) is the kernel of the sign map, and \(V_4\) is the identity together with the entire conjugacy class of double transpositions.
To see there are no others, use the conjugacy classes of \(S_4\), whose sizes are \[ 1,\ 6,\ 3,\ 8,\ 6 \] for cycle types \[ 1^4,\quad 2\,1^2,\quad 2^2,\quad 3\,1,\quad 4, \] respectively. A normal subgroup is a union of conjugacy classes containing the identity, and its order must divide \(24\).
The only such class-size sums that can be orders of proper nontrivial subgroups are \[ 1+3=4 \] and \[ 1+3+8=12. \] These give \(V_4\) and \(A_4\), respectively. All other unions containing the identity have sizes not dividing \(24\), except the full union of size \(24\).
Hence the normal subgroups of \(S_4\) are exactly \[ \{e\},\quad V_4,\quad A_4,\quad S_4, \] so the proper nontrivial ones are precisely \(V_4\) and \(A_4\).