Finite groups containing no subgroups of a given order or index












7














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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  • 5




    $PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
    – YCor
    Nov 24 at 11:37








  • 4




    $PSL_2(11)$ has order $15times 44=660$. 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.
    – YCor
    Nov 24 at 11:50








  • 4




    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 $PSL(3,3)$.
    – M. Farrokhi D. G.
    Nov 24 at 13:39












  • @M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
    – Taras Banakh
    Nov 25 at 7:02










  • @TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
    – YCor
    Nov 25 at 7:50
















7














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?











share|cite|improve this question


















  • 5




    $PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
    – YCor
    Nov 24 at 11:37








  • 4




    $PSL_2(11)$ has order $15times 44=660$. 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.
    – YCor
    Nov 24 at 11:50








  • 4




    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 $PSL(3,3)$.
    – M. Farrokhi D. G.
    Nov 24 at 13:39












  • @M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
    – Taras Banakh
    Nov 25 at 7:02










  • @TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
    – YCor
    Nov 25 at 7:50














7












7








7


1





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?











share|cite|improve this question













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?








gr.group-theory finite-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 10:50









Taras Banakh

15.6k13190




15.6k13190








  • 5




    $PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
    – YCor
    Nov 24 at 11:37








  • 4




    $PSL_2(11)$ has order $15times 44=660$. 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.
    – YCor
    Nov 24 at 11:50








  • 4




    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 $PSL(3,3)$.
    – M. Farrokhi D. G.
    Nov 24 at 13:39












  • @M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
    – Taras Banakh
    Nov 25 at 7:02










  • @TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
    – YCor
    Nov 25 at 7:50














  • 5




    $PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
    – YCor
    Nov 24 at 11:37








  • 4




    $PSL_2(11)$ has order $15times 44=660$. 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.
    – YCor
    Nov 24 at 11:50








  • 4




    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 $PSL(3,3)$.
    – M. Farrokhi D. G.
    Nov 24 at 13:39












  • @M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
    – Taras Banakh
    Nov 25 at 7:02










  • @TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
    – YCor
    Nov 25 at 7:50








5




5




$PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
– YCor
Nov 24 at 11:37






$PSL_2(13)$ has order $21times 52=1092$. 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 $181440$.)
– YCor
Nov 24 at 11:37






4




4




$PSL_2(11)$ has order $15times 44=660$. 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.
– YCor
Nov 24 at 11:50






$PSL_2(11)$ has order $15times 44=660$. 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.
– YCor
Nov 24 at 11:50






4




4




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 $PSL(3,3)$.
– M. Farrokhi D. G.
Nov 24 at 13:39






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 $PSL(3,3)$.
– M. Farrokhi D. G.
Nov 24 at 13:39














@M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
– Taras Banakh
Nov 25 at 7:02




@M.FarrokhiD.G. What is the smallest possible cardinality of a group satisfying the problem?
– Taras Banakh
Nov 25 at 7:02












@TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
– YCor
Nov 25 at 7:50




@TarasBanakh the previous 2 comments gave enough information to answer this question: it's 504. Namely $SL_2(8)$. It has order $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
– YCor
Nov 25 at 7:50










1 Answer
1






active

oldest

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10














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.






share|cite|improve this answer

















  • 1




    ...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.)
    – Ilya Bogdanov
    Nov 24 at 16:38












  • @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.
    – Geoff Robinson
    Nov 24 at 16:44






  • 1




    See here for some non-simple examples.
    – Mikko Korhonen
    Nov 25 at 8:25













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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.






share|cite|improve this answer

















  • 1




    ...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.)
    – Ilya Bogdanov
    Nov 24 at 16:38












  • @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.
    – Geoff Robinson
    Nov 24 at 16:44






  • 1




    See here for some non-simple examples.
    – Mikko Korhonen
    Nov 25 at 8:25


















10














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.






share|cite|improve this answer

















  • 1




    ...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.)
    – Ilya Bogdanov
    Nov 24 at 16:38












  • @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.
    – Geoff Robinson
    Nov 24 at 16:44






  • 1




    See here for some non-simple examples.
    – Mikko Korhonen
    Nov 25 at 8:25
















10












10








10






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 11:11









Geoff Robinson

29.1k278108




29.1k278108








  • 1




    ...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.)
    – Ilya Bogdanov
    Nov 24 at 16:38












  • @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.
    – Geoff Robinson
    Nov 24 at 16:44






  • 1




    See here for some non-simple examples.
    – Mikko Korhonen
    Nov 25 at 8:25
















  • 1




    ...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.)
    – Ilya Bogdanov
    Nov 24 at 16:38












  • @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.
    – Geoff Robinson
    Nov 24 at 16:44






  • 1




    See here for some non-simple examples.
    – Mikko Korhonen
    Nov 25 at 8:25










1




1




...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.)
– Ilya Bogdanov
Nov 24 at 16:38






...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.)
– Ilya Bogdanov
Nov 24 at 16:38














@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.
– Geoff Robinson
Nov 24 at 16:44




@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.
– Geoff Robinson
Nov 24 at 16:44




1




1




See here for some non-simple examples.
– Mikko Korhonen
Nov 25 at 8:25






See here for some non-simple examples.
– Mikko Korhonen
Nov 25 at 8:25




















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