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text: 'If there is a non-trivial [-4h6] $H$ of the [-4hf] $A_5$ on five elements, then [4jw it is a union of conjugacy classes].\nAdditionally, by [4jn Lagrange's theorem], the [3gg order] of a subgroup must divide the order of the group, so the total size of $H$ must divide $60$.\n\nWe can list the [alternating_group_five_conjugacy_classes conjugacy classes of $A_5$]; they are of size $1, 20, 15, 12, 12$ respectively.\n\nBy a brute-force check, no sum of these containing $1$ can possibly divide $60$ (which is the size of $A_5$) unless it is $1$ or $60$.\n\n# The brute-force check\n\nWe first list the [number_theory_divisor divisors] of $60$: they are $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$$\n\nSince the subgroup $H$ must contain the identity, it must contain the conjugacy class of size $1$.\nIf it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $13$ (since the smallest other class is of size $12$); so it is allowed to be one of $15$, $20$, $30$, or $60$.\nSince $H$ is also assumed to be a *proper* subgroup, it cannot be $A_5$ itself, so in fact $60$ is also banned.\n\n## The class of size $20$\nIf $H$ then contains the conjugacy class of size $20$, then $H$ can only be of size $30$ because we have already included $21$ elements.\nBut there is no way to add just $9$ elements using conjugacy classes of size bigger than or equal to $12$.\n\nSo $H$ cannot contain the class of size $20$.\n\n## The class of size $15$\n\nIn this case, $H$ is allowed to be of size $20$ or $30$, and we have already found $16$ elements of it. So there are either $4$ or $14$ elements left to find; but we are only allowed to add classes of size exactly $12$, so this can't be done either.\n\n## The classes of size $12$\n\nWhat remains is two classes of size $12$, from which we can make $1+12 = 13$ or $1+12+12 = 25$.\nNeither of these divides $60$, so these are not legal options either.\n\nThis exhausts the search, and completes the proof.',
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