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text: 'The [-4hf] $A_5$ on five elements is [4jc simple].\n\n# Proof\n\nRecall that $A_5$ has [3gg order] $60$, so [4jn Lagrange's theorem] states that any subgroup of $A_5$ has order dividing $60$.\n\nSuppose $H$ is a normal subgroup of $A_5$, which is not the trivial subgroup $\\{ e \\}$.\nIf $H$ has order divisible by $3$, then by [4l6 Cauchy's theorem] there is a $3$-[49f cycle] in $H$ (because the $3$-cycles are the only elements with order $3$ in $A_5$).\nBecause $H$ [4jw is a union of conjugacy classes], and because the $3$-cycles [4kv form a conjugacy class in $A_n$ for $n > 4$], $H$ would therefore contain *every* $3$-cycle; but then [4hs it would be the entire alternating group].\n\nIf instead $H$ has order divisible by $2$, then there is a double transposition such as $(12)(34)$ in $H$, since these are the only elements of order $2$ in $A_5$.\nBut then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12)(34)$ and $(15)(34)$, so it contains $(15)(34)(12)(34) = (125)$.\nHence as before $H$ contains every $3$-cycle so is the entire alternating group.\n\nSo $H$ must have order exactly $5$, by [4jn Lagrange's theorem]; so it contains an element of order $5$ since [4jh prime order groups are cyclic].\n\nThe only such elements of $A_n$ are $5$-cycles; but the conjugacy class of a $5$-cycle is of size $12$, which is too big to fit in $H$ which has size $5$.',
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