Irreducible factors of primitive permutation group representationpermutation representation of $S_n$Are the distributive permutation groups linearly primitive?Irreducible representation of $C^*(D_infty)$, group $C^*$-algebra of an infinite dihedral groupDecomposition into irreducible components of a representation of $Spin(9)$Regular elementary abelian subgroups of primitive permutation groupscomposition factors of primitive componentsHow do I know if an irreducible representation is a permutation representation?Which groups can be reconstructed from a single invariant subspace?A theory of (or reference for) symmetric point arrangementsConnections between linear representations and permutation representations
Irreducible factors of primitive permutation group representation
permutation representation of $S_n$Are the distributive permutation groups linearly primitive?Irreducible representation of $C^*(D_infty)$, group $C^*$-algebra of an infinite dihedral groupDecomposition into irreducible components of a representation of $Spin(9)$Regular elementary abelian subgroups of primitive permutation groupscomposition factors of primitive componentsHow do I know if an irreducible representation is a permutation representation?Which groups can be reconstructed from a single invariant subspace?A theory of (or reference for) symmetric point arrangementsConnections between linear representations and permutation representations
$begingroup$
Consider a primitive permutation group $GammasubseteqmathrmSym(N)$ on the $n$-element set $N=1,...,n$, that is, $Gamma$ does not preserve any non-trivial partition of $N$.
Consider the linear representation of $Gamma$ by permutation matrices on $Bbb R^n$ in the obvious way.
What can be said about the decomposition of this representation into irreducible factors?
I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic?
gr.group-theory rt.representation-theory permutation-groups
$endgroup$
add a comment |
$begingroup$
Consider a primitive permutation group $GammasubseteqmathrmSym(N)$ on the $n$-element set $N=1,...,n$, that is, $Gamma$ does not preserve any non-trivial partition of $N$.
Consider the linear representation of $Gamma$ by permutation matrices on $Bbb R^n$ in the obvious way.
What can be said about the decomposition of this representation into irreducible factors?
I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic?
gr.group-theory rt.representation-theory permutation-groups
$endgroup$
3
$begingroup$
There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
$endgroup$
– Richard Lyons
Jul 29 at 14:44
add a comment |
$begingroup$
Consider a primitive permutation group $GammasubseteqmathrmSym(N)$ on the $n$-element set $N=1,...,n$, that is, $Gamma$ does not preserve any non-trivial partition of $N$.
Consider the linear representation of $Gamma$ by permutation matrices on $Bbb R^n$ in the obvious way.
What can be said about the decomposition of this representation into irreducible factors?
I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic?
gr.group-theory rt.representation-theory permutation-groups
$endgroup$
Consider a primitive permutation group $GammasubseteqmathrmSym(N)$ on the $n$-element set $N=1,...,n$, that is, $Gamma$ does not preserve any non-trivial partition of $N$.
Consider the linear representation of $Gamma$ by permutation matrices on $Bbb R^n$ in the obvious way.
What can be said about the decomposition of this representation into irreducible factors?
I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic?
gr.group-theory rt.representation-theory permutation-groups
gr.group-theory rt.representation-theory permutation-groups
asked Jul 29 at 12:59
M. WinterM. Winter
1,4767 silver badges23 bronze badges
1,4767 silver badges23 bronze badges
3
$begingroup$
There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
$endgroup$
– Richard Lyons
Jul 29 at 14:44
add a comment |
3
$begingroup$
There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
$endgroup$
– Richard Lyons
Jul 29 at 14:44
3
3
$begingroup$
There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
$endgroup$
– Richard Lyons
Jul 29 at 14:44
$begingroup$
There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
$endgroup$
– Richard Lyons
Jul 29 at 14:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An example in which there are two isomorphic irreducible modules in the decomposition is the group $rm PSL(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.
Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
$endgroup$
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
add a comment |
$begingroup$
Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $fracchi(1)m.$
Recall that a base for the permutation group $G$ acting on $Omega$ is a subset $beta$ of $Omega$ such that only the identity element of $G$ fixes every element of $beta$.
Hence we obtain $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$
$endgroup$
add a comment |
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$begingroup$
An example in which there are two isomorphic irreducible modules in the decomposition is the group $rm PSL(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.
Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
$endgroup$
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
add a comment |
$begingroup$
An example in which there are two isomorphic irreducible modules in the decomposition is the group $rm PSL(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.
Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
$endgroup$
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
add a comment |
$begingroup$
An example in which there are two isomorphic irreducible modules in the decomposition is the group $rm PSL(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.
Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
$endgroup$
An example in which there are two isomorphic irreducible modules in the decomposition is the group $rm PSL(2,11)$ in its primitive permutation representation of degree $55$ coming from the action of $G$ on the cosets of a dihedral subgroup of order $12$. The permutation module over the real numbers decomposes into modules of dimensions $1,10,10,10,12,12$, where two of the $10$-dimensional constituents are isomorphic.
Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $1+5+5+10+10+12+12$, but note that the two $5$-dimensional constituents are contragredient, and they combine to make a $10$- real representation.
> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
| A(1, 11) = L(2, 11)
1
> CT := CharacterTable(G);
> CT;
Character Table of Group G
--------------------------
-------------------------------------------
Class | 1 2 3 4 5 6 7 8
Size | 1 55 110 132 132 110 60 60
Order | 1 2 3 5 5 6 11 11
-------------------------------------------
p = 2 1 1 3 5 4 3 8 7
p = 3 1 2 1 5 4 2 7 8
p = 5 1 2 3 1 1 6 7 8
p = 11 1 2 3 4 5 6 1 1
-------------------------------------------
X.1 + 1 1 1 1 1 1 1 1
X.2 0 5 1 -1 0 0 1 Z2 Z2#2
X.3 0 5 1 -1 0 0 1 Z2#2 Z2
X.4 + 10 -2 1 0 0 1 -1 -1
X.5 + 10 2 1 0 0 -1 -1 -1
X.6 + 11 -1 -1 1 1 -1 0 0
X.7 + 12 0 0 Z1 Z1#2 0 1 1
X.8 + 12 0 0 Z1#2 Z1 0 1 1
Explanation of Character Value Symbols
# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k
Z1 = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2 = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]
> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]
answered Jul 29 at 14:00
Derek HoltDerek Holt
28.3k4 gold badges66 silver badges115 bronze badges
28.3k4 gold badges66 silver badges115 bronze badges
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
add a comment |
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
$begingroup$
Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one.
$endgroup$
– M. Winter
Jul 29 at 15:58
2
2
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
$begingroup$
Yes, the primitive permutation representation of degree $91$ of $rm PSL(2,13)$ has a repeated real constituent of degree $13$.
$endgroup$
– Derek Holt
Jul 30 at 12:18
add a comment |
$begingroup$
Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $fracchi(1)m.$
Recall that a base for the permutation group $G$ acting on $Omega$ is a subset $beta$ of $Omega$ such that only the identity element of $G$ fixes every element of $beta$.
Hence we obtain $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$
$endgroup$
add a comment |
$begingroup$
Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $fracchi(1)m.$
Recall that a base for the permutation group $G$ acting on $Omega$ is a subset $beta$ of $Omega$ such that only the identity element of $G$ fixes every element of $beta$.
Hence we obtain $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$
$endgroup$
add a comment |
$begingroup$
Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $fracchi(1)m.$
Recall that a base for the permutation group $G$ acting on $Omega$ is a subset $beta$ of $Omega$ such that only the identity element of $G$ fixes every element of $beta$.
Hence we obtain $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$
$endgroup$
Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $G$ is a (faithful) primitive permutation group of degree $n$, and a complex irreducible character $chi$ of $G$ occurs with multiplicity $m$ in the associated permutation character of degree $n$, then there is a base for $G$ of size at most $fracchi(1)m.$
Recall that a base for the permutation group $G$ acting on $Omega$ is a subset $beta$ of $Omega$ such that only the identity element of $G$ fixes every element of $beta$.
Hence we obtain $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$
answered Jul 29 at 14:34
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
30.8k2 gold badges85 silver badges117 bronze badges
add a comment |
add a comment |
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There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_24$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$.
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– Richard Lyons
Jul 29 at 14:44