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  "body": "\n## Proof\n\nWe show that a nontrivial normal subgroup contains a $3$-cycle, and then all $3$-cycles. Products in $S_n$ are composites of maps, and the right factor acts first.\n\n1. Let $n \\geq 5$ and let $N$ be a [[normal subgroup|0W77CVBCHQ]] of $A_n$ with $N \\neq \\{e\\}$. We must show $N = A_n$.\n2. Conjugation formula. Let $\\rho \\in S_n$ and let $c = (b_1\\ \\dots\\ b_k)$ be a cycle in [[cycle notation|0ZKPH2BD1T]]. For each $i$, $\\rho c \\rho^{-1}(\\rho(b_i)) = \\rho(c(b_i))$, which is $\\rho(b_{i+1})$, with $b_{k+1} = b_1$.\n3. If $y$ is not among the $\\rho(b_i)$, then $\\rho^{-1}(y)$ is not among the $b_i$, so $c$ fixes it and $\\rho c \\rho^{-1}(y) = y$. Hence $\\rho c \\rho^{-1} = (\\rho(b_1)\\ \\dots\\ \\rho(b_k))$.\n4. For a product $c_1 \\cdots c_r$, inserting $\\rho^{-1} \\rho$ between the factors gives $\\rho c_1 \\cdots c_r \\rho^{-1} = (\\rho c_1 \\rho^{-1}) \\cdots (\\rho c_r \\rho^{-1})$. If $\\rho$ fixes every point moved by $c$, step 3 gives $\\rho c \\rho^{-1} = c$.\n5. All $3$-cycles are conjugate in $A_n$. Let $(a\\ b\\ c)$ and $(a'\\ b'\\ c')$ be $3$-cycles. Choose $\\rho \\in S_n$ with $\\rho(a) = a'$, $\\rho(b) = b'$, $\\rho(c) = c'$.\n6. Such $\\rho$ exists, because the complements of $\\{a, b, c\\}$ and $\\{a', b', c'\\}$ both have $n - 3$ elements, and any bijection between them extends the choice. By step 3, $\\rho (a\\ b\\ c) \\rho^{-1} = (a'\\ b'\\ c')$.\n7. If $\\rho$ is odd, choose distinct points $d, e$ outside $\\{a, b, c\\}$, which is possible since $n \\geq 5$. Put $\\rho' = \\rho (d\\ e)$.\n8. By the [[sign homomorphism|0JEJJCB0EX]], $\\rho'$ is even, so $\\rho' \\in A_n$. [[Disjoint cycles commute|07E6D8R4D6]], so $(d\\ e)(a\\ b\\ c)(d\\ e)^{-1} = (a\\ b\\ c)$, and $\\rho' (a\\ b\\ c) \\rho'^{-1} = (a'\\ b'\\ c')$.\n9. Hence, if $N$ contains one $3$-cycle, normality gives all $3$-cycles in $N$. Since the [[$3$-cycles generate $A_n$|0KFZ0QX83G]], the [[generated subgroup|0RG9SPQSWM]] lies in $N$, so $N = A_n$.\n10. It remains to find a $3$-cycle in $N$. Key fact: for $\\sigma \\in N$ and $\\rho \\in A_n$, normality gives $\\rho \\sigma \\rho^{-1} \\in N$. So $\\pi = (\\rho \\sigma \\rho^{-1}) \\sigma^{-1} \\in N$.\n11. Pick $\\sigma \\in N$ with $\\sigma \\neq e$. By the [[disjoint cycle decomposition|0YWDAZPRPQ]], $\\sigma$ is a product of disjoint cycles of length at least $2$, and there is at least one cycle.\n12. In the cases below, $\\sigma = c \\tau$. Here $c$ is a product of listed cycles, which move the points of a set $S$, and $\\tau$ is the product of the other cycles. Each $\\rho$ used moves only points of $S$.\n13. Then $\\rho \\sigma \\rho^{-1} = (\\rho c \\rho^{-1}) \\tau$ by step 4, and $\\rho c \\rho^{-1}$ moves only points of $S$. For $x \\notin S$, $\\sigma^{-1}(x) = \\tau^{-1}(x) \\notin S$, so $\\pi(x) = \\tau(\\tau^{-1}(x)) = x$. So we only compute $\\pi$ on $S$.\n14. Case A: some cycle has length $r \\geq 4$, say $\\sigma = (a_1\\ a_2\\ \\dots\\ a_r) \\tau$. Take $\\rho = (a_1\\ a_2\\ a_3)$, which is even by the [[sign of a cycle|0V0E04DWX2]].\n15. By step 3, $\\rho \\sigma \\rho^{-1} = (a_2\\ a_3\\ a_1\\ a_4\\ \\dots\\ a_r) \\tau$. On $S$, $\\sigma^{-1}$ sends $a_1 \\mapsto a_r$ and $a_j \\mapsto a_{j-1}$ for $j \\geq 2$.\n16. So $\\pi(a_1) = a_2$, $\\pi(a_2) = a_4$, $\\pi(a_3) = a_3$, $\\pi(a_4) = a_1$, and $\\pi(a_j) = a_j$ for $j \\geq 5$. Thus $\\pi = (a_1\\ a_2\\ a_4)$ is a $3$-cycle in $N$.\n17. Case B: no cycle has length $4$ or more, and there are at least two $3$-cycles, say $\\sigma = (a_1\\ a_2\\ a_3)(a_4\\ a_5\\ a_6) \\tau$. Take $\\rho = (a_1\\ a_2\\ a_4) \\in A_n$.\n18. By step 3, $\\rho \\sigma \\rho^{-1} = (a_2\\ a_4\\ a_3)(a_1\\ a_5\\ a_6) \\tau$. Computing on $S$ gives $\\pi(a_1) = a_2$, $\\pi(a_2) = a_5$, $\\pi(a_5) = a_3$, $\\pi(a_3) = a_4$, $\\pi(a_4) = a_1$, $\\pi(a_6) = a_6$.\n19. So $\\pi = (a_1\\ a_2\\ a_5\\ a_3\\ a_4)$ lies in $N$. Applying Case A to $\\pi$ in place of $\\sigma$ gives a $3$-cycle in $N$.\n20. Case C: no cycle has length $4$ or more, and there is exactly one $3$-cycle. So $\\sigma = (a_1\\ a_2\\ a_3) \\tau$, where $\\tau$ is a product of disjoint transpositions, possibly empty.\n21. The factors commute, since [[disjoint cycles commute|07E6D8R4D6]], and each transposition squares to $e$. So $\\sigma^2 = (a_1\\ a_2\\ a_3)^2 \\tau^2 = (a_1\\ a_3\\ a_2)$, a $3$-cycle in $N$.\n22. Case D: every cycle is a transposition. If there are $m$ of them, the [[definition of the sign|0SK90238BH]] gives $(-1)^m = 1$, because $\\sigma \\in A_n$ by the [[definition of $A_n$|0T13HM377F]].\n23. So $m$ is even, and $m \\geq 2$ since $\\sigma \\neq e$. Write $\\sigma = (a_1\\ a_2)(a_3\\ a_4) \\tau$ and take $\\rho = (a_1\\ a_2\\ a_3) \\in A_n$.\n24. By step 3, $\\rho \\sigma \\rho^{-1} = (a_2\\ a_3)(a_1\\ a_4) \\tau$. Since $\\sigma^{-1} = \\sigma$, computing on $S$ gives $\\pi = (a_1\\ a_3)(a_2\\ a_4) \\in N$.\n25. Since $n \\geq 5$, choose $a_5 \\notin \\{a_1, a_2, a_3, a_4\\}$ and let $\\beta = (a_1\\ a_3\\ a_5) \\in A_n$. By step 3, $\\beta \\pi \\beta^{-1} = (a_3\\ a_5)(a_2\\ a_4) \\in N$.\n26. Then $\\pi \\beta \\pi \\beta^{-1} = (a_1\\ a_3)(a_2\\ a_4)(a_3\\ a_5)(a_2\\ a_4)$ lies in $N$. Swapping the disjoint factors $(a_3\\ a_5)$ and $(a_2\\ a_4)$ and cancelling $(a_2\\ a_4)(a_2\\ a_4) = e$ makes it $(a_1\\ a_3)(a_3\\ a_5)$.\n27. Point by point, $(a_1\\ a_3)(a_3\\ a_5)$ sends $a_1 \\mapsto a_3$, $a_3 \\mapsto a_5$, $a_5 \\mapsto a_1$. So it is the $3$-cycle $(a_1\\ a_3\\ a_5)$, and it lies in $N$.\n28. Cases A to D cover every $\\sigma \\neq e$. Either some cycle has length at least $4$, or all lengths are $2$ or $3$. In the second case, the number of $3$-cycles is at least two, exactly one, or zero.\n29. So $N$ contains a $3$-cycle, and $N = A_n$ by step 9. Hence the only normal subgroups of $A_n$ are $\\{e\\}$ and $A_n$.\n\n## Idea\n\nFor $n \\geq 5$ there is room to correct the parity of a conjugating permutation, so all $3$-cycles are conjugate inside $A_n$. A normal subgroup is closed under the commutator-like products $\\rho \\sigma \\rho^{-1} \\sigma^{-1}$. A good choice of the $3$-cycle $\\rho$ makes this product a $3$-cycle, which then forces the whole group.\n",
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