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?)
finite soluble groups and Hall subgroups
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finite-groups
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0Are you asking how you know what $A_5$ has no subgroups of index $4$? – 2012-03-06
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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.