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It is rather obvious that every Hall π-subgroup is a Sylow π-subgroup. In general, however, G need not contain any Hall π-subgroups. For example, a Hall {3,5}-subgroup of $A_5$ would have index 4, but $A_5$ have no such subgroups. (Why?)

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Are you asking how you know what $A_5$ has no subgroups of index $4$? –  Arturo Magidin Mar 6 '12 at 16:27

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The converse of Hall's Theorem says that a group $G$ must be solvable if it has Hall $\pi$-subgroups for all sets $\pi$ of prime divisors of $|G|$. Your example, $A_5$, is not solvable, so by this theorem there must be sets of primes (such as the set you found) with no corresponding Hall subgroup.

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