Show that \(S_n\) for \(n \ge 5\) has exactly one non-trivial proper normal subgroup: \(A_n\).
Solution
For \(n\ge5\), \(A_n\) is simple (E-AMD-6TLQZZWT).
Let \(N\trianglelefteq S_n\) be nontrivial. Then \[ N\cap A_n\trianglelefteq A_n. \] By simplicity of \(A_n\), either \[ N\cap A_n=A_n \qquad\text{or}\qquad N\cap A_n=1. \]
1. If \(N\cap A_n=A_n\), then \(N=A_n\) or \(N=S_n\).
In this case \(A_n\le N\le S_n\). Since \([S_n:A_n]=2\), there is no intermediate subgroup strictly between them.
2. The case \(N\cap A_n=1\) is impossible.
The quotient map \(S_n\to S_n/A_n\cong C_2\) restricts injectively to \(N\), because its kernel on \(N\) is \(N\cap A_n=1\). Since \(N\ne1\), it follows that \(|N|=2\).
Write \(N=\{1,\sigma\}\). Normality implies \[ g\sigma g^{-1}\in N\setminus\{1\}=\{\sigma\} \] for every \(g\in S_n\). Hence \(\sigma\in Z(S_n)\). But \(Z(S_n)=1\) for \(n\ge3\), contradiction.
Thus every nontrivial proper normal subgroup of \(S_n\) equals \(A_n\). Therefore \[ \boxed{A_n\text{ is the unique nontrivial proper normal subgroup of }S_n\quad(n\ge5).} \]