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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?

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    $\begingroup$ $PSL_2(13)$ has order $21\times 52=1092$. According to Corollary 2.2 in staff.ncl.ac.uk/o.h.king/KingBCC05.pdf, its maximal subgroups have order $12$, $14$ and $78$. Hence it has no subgroup of order or index $21$ (i.e. order $21$ or $52$), which can certainly be checked by hand. (Note that $A_9$ has cardinal $181440$.) $\endgroup$ – YCor Nov 24 '18 at 11:37
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    $\begingroup$ $PSL_2(11)$ has order $15\times 44=660$. Using the same reference, its maximal subgroups have order $12,55,60$, those of order $60$ being alternating groups and thus not having subgroups of order $15$. Hence $PSL_2(11)$ has no subgroup of order or index $15$. There are only 4 smaller nonabelian finite simple groups, of order $60$, $168$, $360$, $504$. The first 3 give no example. $\endgroup$ – YCor Nov 24 '18 at 11:50
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    $\begingroup$ A simple verification by GAP shows that the only non-abelian simple groups of orders less than $1,000,000$ which do not satisfy the problem are $A_5$, $A_6$, $PSL(3,2)$, and $PSL(3,3)$. $\endgroup$ – M. Farrokhi D. G. Nov 24 '18 at 13:39
  • $\begingroup$ @M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem? $\endgroup$ – Taras Banakh Nov 25 '18 at 7:02
  • $\begingroup$ @TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21\times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$. $\endgroup$ – YCor Nov 25 '18 at 7:50
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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.

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    $\begingroup$ ...or to check the products of Sylow 2- and 3-subgroups. (likewise, one may check that there is no product of 5- and 7-Sylow that is a subgroup.) $\endgroup$ – Ilya Bogdanov Nov 24 '18 at 16:38
  • $\begingroup$ @IlyaBogdanov :Yes, there is no subgroup of order $35$ as an easy application of Sylow's Theorem shows that such a subgroup would have to be Abelian ( whereas $A_{9}$ has a self-centralizing Sylow $7$-subgroup). Also, in general, it is very rare for $A_{n}$ to have Hall subgroups which are not Sylow subgroups. $\endgroup$ – Geoff Robinson Nov 24 '18 at 16:44
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    $\begingroup$ See here for some non-simple examples. $\endgroup$ – Mikko Korhonen Nov 25 '18 at 8:25

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