Finite groups containing no subgroups of a given order or index
$begingroup$
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
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
$endgroup$
|
show 3 more comments
$begingroup$
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
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
$endgroup$
5
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
4
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
4
$begingroup$
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 $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50
|
show 3 more comments
$begingroup$
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
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
$endgroup$
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
gr.group-theory finite-groups
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
asked Nov 24 '18 at 10:50
Taras BanakhTaras Banakh
16.9k13495
16.9k13495
5
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
4
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
4
$begingroup$
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 $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50
|
show 3 more comments
5
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
4
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
4
$begingroup$
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 $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50
5
5
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
4
4
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
4
4
$begingroup$
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
$begingroup$
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
$begingroup$
@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 $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50
$begingroup$
@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 $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50
|
show 3 more comments
1 Answer
1
active
oldest
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$begingroup$
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.
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
$endgroup$
1
$begingroup$
...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
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
add a comment |
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$begingroup$
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.
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
$endgroup$
1
$begingroup$
...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
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
$endgroup$
1
$begingroup$
...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
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
$endgroup$
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.
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
answered Nov 24 '18 at 11:11
Geoff RobinsonGeoff Robinson
29.5k280109
29.5k280109
1
$begingroup$
...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
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
add a comment |
1
$begingroup$
...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
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
1
1
$begingroup$
...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
$begingroup$
...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
$begingroup$
@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
1
1
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
$begingroup$
See here for some non-simple examples.
$endgroup$
– Mikko Korhonen
Nov 25 '18 at 8:25
add a comment |
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5
$begingroup$
$PSL_2(13)$ 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 $181440$.)
$endgroup$
– YCor
Nov 24 '18 at 11:37
4
$begingroup$
$PSL_2(11)$ 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.
$endgroup$
– YCor
Nov 24 '18 at 11:50
4
$begingroup$
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 $21times 24$ and its maximal subgroups have order $14$, $18$, $56$, $6$.
$endgroup$
– YCor
Nov 25 '18 at 7:50