- Show that \(S_{n\geq 5}\) has one normal subgroup: \(A_n\).
Solution
For \(n\ge 5\), the normal subgroups of \(S_n\) are \(1\), \(A_n\) and \(S_n\), so \(A_n\) is its unique proper nontrivial normal subgroup.
1. For \(n\ge 5\), \(A_n\) is simple.
This is E-AMD-6TLQZZWT.
2. Let \(N\trianglelefteq S_n\).Then \(N\cap A_n\trianglelefteq A_n\), so\[ N\cap A_n\in\{1,A_n\}. \]
The intersection of two normal subgroups of \(S_n\) is normal in \(S_n\), hence in \(A_n\). Apply 1.
3. If \(N\cap A_n=A_n\), then either \(N=A_n\) or \(N=S_n\).
In this case \(A_n\le N\le S_n\). Since \([S_n:A_n]=2\), there are no intermediate subgroups other than the endpoints.
4. If \(N\cap A_n=1\), then \(N=1\).
The composite \[ N\hookrightarrow S_n\twoheadrightarrow S_n/A_n\cong C_2 \] has kernel \(N\cap A_n=1\), so \(N\) embeds in \(C_2\). Thus \(|N|\le 2\). If \(|N|=2\), write \(N=\{1,\tau\}\). Normality forces \(g\tau g^{-1}=\tau\) for every \(g\in S_n\), so \(\tau\in Z(S_n)\). But \(Z(S_n)=1\) for \(n\ge 3\): if a nonidentity permutation \(\sigma\) moves \(i\) to \(j\ne i\), choose \(k\notin\{i,j\}\); then the transposition \((j\ k)\) does not commute with \(\sigma\). Hence \(|N|\ne2\), so \(N=1\).
5. Therefore the normal subgroups of \(S_n\) are exactly\[ 1,\qquad A_n,\qquad S_n, \]and \(A_n\) is the unique proper nontrivial normal subgroup.
Combine <1>2–<1>4.