Show that \(S_4\) has two normal subgroups: \(A_4, \ZZ_2^2\).
Solution
1. The alternating group \(A_4\) is normal in \(S_4\).
The sign map \[ \operatorname{sgn}:S_4\longrightarrow\{\pm1\} \] is a group homomorphism, and \(A_4=\ker(\operatorname{sgn})\). Every kernel is normal, so \(A_4\trianglelefteq S_4\).
2. The set\[ V=\{1,(12)(34),(13)(24),(14)(23)\} \]is a subgroup isomorphic to \(\ZZ_2^2\).
Each nonidentity element of \(V\) has order \(2\), and the product of any two distinct nonidentity elements is the third; for example, \[ (12)(34)(13)(24)=(14)(23). \] Thus \(V\) is closed under products and inverses, so \(V\le S_4\). Since \(V\) has order \(4\) and every nonidentity element has order \(2\), it is isomorphic to \(\ZZ_2^2\).
3. The subgroup \(V\) is normal in \(S_4\).
For \(\sigma\in S_4\) and a double transposition \((ab)(cd)\), \[ \sigma(ab)(cd)\sigma^{-1} =(\sigma(a)\,\sigma(b))(\sigma(c)\,\sigma(d)), \] which is again a double transposition. There are exactly three double transpositions in \(S_4\), namely the three nonidentity elements of \(V\). Hence conjugation by every \(\sigma\in S_4\) preserves \(V\), so \(V\trianglelefteq S_4\).
4. Therefore \(S_4\) has the two stated normal subgroups \(A_4\) and \(\ZZ_2^2\).
Combine <1>1–<1>3, identifying \(V\cong\ZZ_2^2\).