- Argue that \(A_n\) is simple for \(n \geq 5\).
Solution
Let \(n\ge5\), and let \(N\trianglelefteq A_n\) be nontrivial. We prove \(N=A_n\). For a permutation \(\sigma\), write \(\operatorname{supp}(\sigma)\) for the set of points moved by \(\sigma\).
1. Choose \(1\ne\sigma\in N\) with \(|\operatorname{supp}(\sigma)|\) minimal. Then \(|\operatorname{supp}(\sigma)|\le5\).
Suppose instead that \(\sigma\) moves at least six points. Choose a moved point \(a\) and put \(b=\sigma(a)\).
If \(\sigma(b)\notin\{a,b\}\), choose \[ c\notin\{a,b,\sigma(b)\}. \] If \(\sigma(b)=a\), so that \((a\ b)\) is a \(2\)-cycle of \(\sigma\), choose a moved point \(c\) outside \(\{a,b\}\); then \(\sigma(c)\neq c\), and \(\sigma(c)\notin\{a,b\}\) because \(\sigma^{-1}(a)=b\) and \(\sigma^{-1}(b)=a\), so \(\sigma(c)\notin\{a,b,c\}\).
Set \[ \tau=(a\ b\ c)\in A_n \] and consider \[ \rho=\tau\sigma\tau^{-1}\sigma^{-1}\in N. \] The inclusion \(\rho\in N\) follows from normality. Moreover, \[ \operatorname{supp}(\rho) \subseteq \operatorname{supp}(\tau)\cup \sigma(\operatorname{supp}(\tau)) =\{a,b,c\}\cup\{b,\sigma(b),\sigma(c)\}, \] so \(\rho\) moves at most five points.
By our choice of \(c\), the set \(\{a,b,c\}\) is not preserved by \(\sigma\). Hence \(\sigma\tau\sigma^{-1}\ne\tau\), so \(\rho\ne1\). Thus \(N\) contains a nonidentity element moving at most five points, contradicting the minimality of \(\sigma\). Therefore \(|\operatorname{supp}(\sigma)|\le5\).
2. The subgroup \(N\) contains a \(3\)-cycle.
A nonidentity even permutation cannot move exactly two points. If the minimal-support element \(\sigma\) moves three points, then it is a \(3\)-cycle and we are done.
If \(\sigma\) moves four points, then, being even and having no fixed point on its support, it is a product of two disjoint transpositions. After relabeling, \[ \sigma=(1\ 2)(3\ 4). \] Since \(n\ge5\), let \(\tau=(1\ 2\ 5)\). Direct calculation gives \[ \tau\sigma\tau^{-1}\sigma^{-1}=(1\ 5\ 2), \] a \(3\)-cycle in \(N\).
If \(\sigma\) moves five points, the only fixed-point-free even cycle type on five points is a \(5\)-cycle. After relabeling, \[ \sigma=(1\ 2\ 3\ 4\ 5). \] Taking \(\tau=(1\ 2\ 3)\) gives \[ \tau\sigma\tau^{-1}\sigma^{-1}=(1\ 2\ 4), \] Again this is a \(3\)-cycle in \(N\).
Thus in every case \(N\) contains a \(3\)-cycle.
3. All \(3\)-cycles are conjugate in \(A_n\) for \(n\ge5\).
Any two \(3\)-cycles are conjugate in \(S_n\). Let \(\alpha\) and \(\beta\) be \(3\)-cycles and choose \(g\in S_n\) with \(g\alpha g^{-1}=\beta\). If \(g\) is even, there is nothing to prove. If \(g\) is odd, choose two points outside the support of \(\alpha\); this is possible because \(n\ge5\). Let \(t\) be the transposition of those two points. Then \(t\) commutes with \(\alpha\), and \(gt\) is even. Hence \[ (gt)\alpha(gt)^{-1}=g\alpha g^{-1}=\beta. \] So the conjugating element may be chosen in \(A_n\).
4. The \(3\)-cycles generate \(A_n\).
Every element of \(A_n\) is a product of an even number of transpositions. Pair the transpositions. If two transpositions share a point, their product is a \(3\)-cycle (or the identity). If they are disjoint, then \[ (a\ b)(c\ d)=(a\ c\ b)(a\ c\ d), \] a product of two \(3\)-cycles. Hence every even permutation is a product of \(3\)-cycles.
5. Therefore \(N=A_n\), so \(A_n\) is simple.
By 2, \(N\) contains one \(3\)-cycle. Since \(N\) is normal, 3 implies that it contains every \(3\)-cycle. By 4, those elements generate \(A_n\). Hence \(N=A_n\).
Thus the only normal subgroups of \(A_n\) are \(1\) and \(A_n\) for every \(n\ge5\).