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.



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> **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?
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Finite groups containing no subgroups of a given order or index
The alternating group $A\_{9}$ has no subgroup of order $35$ and no subgroup of index $35$. This can be checked from the Atlas ( or probably with GAP, etc). Note that if there were a subgroup of index $35,$ it would have to be maximal and its order would be $2^{6}.3^{4}$, so it is only necessary to check maximal subgroups, and there are none of that order.


