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  "body": "\n## Statement\n\nLet $n \\geq 5$.\nThen the [[alternating group|0Z5905Q73D]] $A_n$ is simple: its only [[normal subgroups|0CB0DX5BVV]] are $\\{e\\}$ and $A_n$.\n\n## Remarks\n\nA usual proof shows that a nontrivial normal subgroup of $A_n$ contains a $3$-cycle, then all $3$-cycles, and then uses that the [[3-cycles generate|0KFZ0QX83G]] $A_n$.\nThe group $A_4$ is not simple, because the Klein four-group $\\{e, (1\\ 2)(3\\ 4), (1\\ 3)(2\\ 4), (1\\ 4)(2\\ 3)\\}$ is normal in $A_4$.\nThe group $A_3$ is simple because it has prime order $3$.\nFor $n \\geq 5$, the only normal subgroups of $S_n$ are $\\{e\\}$, $A_n$ and $S_n$.\n\n## Explanations\n\n### L1 human\n\nA simple group cannot be broken into a normal subgroup and a quotient.\nFrom five symbols on, the even permutations form such an unbreakable group.\nThis fact is a key step in the proof that there is no general formula in radicals for polynomials of degree $5$ or more.\n",
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