The classical Lagrange's Theorem says that the order of any subgroup of a finite group divides the order of the group. For abelian groups this theorem can be completed by the following simple fact: Abelian groups contain subgroups of any order that divides the order of the group. For non-abelian groups this is not true: the alternating group $A_4$ has cardinality 12 but contains no subgroup of cardinality 6; the alternating group $A_5$ has order 60 but contains no subgroup of order 30. On the other hand, these alternating groups contain subgroups of index 6 and 30, respectively.
Problem. Is there a finite group $G$ and a divisor $d$ of $|G|$ such that $G$ contains no subgroup of order or index, equal to $d$? Is there such a group $G$ among finite simple groups?