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













5












$begingroup$


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?










share|cite|improve this question









$endgroup$









  • 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















5












$begingroup$


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?










share|cite|improve this question









$endgroup$









  • 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













5












5








5





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










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












  • 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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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 ]





share|cite|improve this answer









$endgroup$














  • $begingroup$
    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


















3












$begingroup$

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 $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$






share|cite|improve this answer









$endgroup$

















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    2 Answers
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    4












    $begingroup$

    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 ]





    share|cite|improve this answer









    $endgroup$














    • $begingroup$
      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















    4












    $begingroup$

    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 ]





    share|cite|improve this answer









    $endgroup$














    • $begingroup$
      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













    4












    4








    4





    $begingroup$

    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 ]





    share|cite|improve this answer









    $endgroup$



    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 ]






    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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
















    • $begingroup$
      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











    3












    $begingroup$

    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 $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$






    share|cite|improve this answer









    $endgroup$



















      3












      $begingroup$

      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 $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$






      share|cite|improve this answer









      $endgroup$

















        3












        3








        3





        $begingroup$

        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 $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$






        share|cite|improve this answer









        $endgroup$



        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 $|G| leq n(n-1) ldots (n+1 - fracchi(1)m) < n^fracchi(1)m.$







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        answered Jul 29 at 14:34









        Geoff RobinsonGeoff Robinson

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