Which partitions realise group algebras of finite groups?











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Fix an algebraically closed field $K$ (maybe of characteristic zero first for simplicity, like $mathbb{C}$).
Given a partition $p=[a_1,...,a_m]$ of an integer $n$. We can identify $p$ with the semisimple algebra $A_p:=M_{a_1}(K) times cdots times M_{a_m}(K)$.



Questions:




  1. Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?


  2. Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?



  3. How does the sequence of numbers of such partitions begin depending on $K$ (Does it in general depend on the field or perhaps just the characteristic of the field?)?



    Probably the answer is very complicated, but can something be said about the rough behavior of the sequence?




It should start as follows for $n geq 1$ and $K=mathbb{C}$, using GAP: 1,1,1,2,1,4,1,4,2,4,1,9,1,5,4





  1. $textbf{Are those numbers always powers of primes?}$. (The next term takes forever to calculate, but maybe I should wait longer before asking this question....)


I dont know how complicated those questions are, so partial answers are also welcome.










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    up vote
    12
    down vote

    favorite
    3












    Fix an algebraically closed field $K$ (maybe of characteristic zero first for simplicity, like $mathbb{C}$).
    Given a partition $p=[a_1,...,a_m]$ of an integer $n$. We can identify $p$ with the semisimple algebra $A_p:=M_{a_1}(K) times cdots times M_{a_m}(K)$.



    Questions:




    1. Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?


    2. Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?



    3. How does the sequence of numbers of such partitions begin depending on $K$ (Does it in general depend on the field or perhaps just the characteristic of the field?)?



      Probably the answer is very complicated, but can something be said about the rough behavior of the sequence?




    It should start as follows for $n geq 1$ and $K=mathbb{C}$, using GAP: 1,1,1,2,1,4,1,4,2,4,1,9,1,5,4





    1. $textbf{Are those numbers always powers of primes?}$. (The next term takes forever to calculate, but maybe I should wait longer before asking this question....)


    I dont know how complicated those questions are, so partial answers are also welcome.










    share|cite|improve this question


























      up vote
      12
      down vote

      favorite
      3









      up vote
      12
      down vote

      favorite
      3






      3





      Fix an algebraically closed field $K$ (maybe of characteristic zero first for simplicity, like $mathbb{C}$).
      Given a partition $p=[a_1,...,a_m]$ of an integer $n$. We can identify $p$ with the semisimple algebra $A_p:=M_{a_1}(K) times cdots times M_{a_m}(K)$.



      Questions:




      1. Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?


      2. Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?



      3. How does the sequence of numbers of such partitions begin depending on $K$ (Does it in general depend on the field or perhaps just the characteristic of the field?)?



        Probably the answer is very complicated, but can something be said about the rough behavior of the sequence?




      It should start as follows for $n geq 1$ and $K=mathbb{C}$, using GAP: 1,1,1,2,1,4,1,4,2,4,1,9,1,5,4





      1. $textbf{Are those numbers always powers of primes?}$. (The next term takes forever to calculate, but maybe I should wait longer before asking this question....)


      I dont know how complicated those questions are, so partial answers are also welcome.










      share|cite|improve this question















      Fix an algebraically closed field $K$ (maybe of characteristic zero first for simplicity, like $mathbb{C}$).
      Given a partition $p=[a_1,...,a_m]$ of an integer $n$. We can identify $p$ with the semisimple algebra $A_p:=M_{a_1}(K) times cdots times M_{a_m}(K)$.



      Questions:




      1. Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?


      2. Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?



      3. How does the sequence of numbers of such partitions begin depending on $K$ (Does it in general depend on the field or perhaps just the characteristic of the field?)?



        Probably the answer is very complicated, but can something be said about the rough behavior of the sequence?




      It should start as follows for $n geq 1$ and $K=mathbb{C}$, using GAP: 1,1,1,2,1,4,1,4,2,4,1,9,1,5,4





      1. $textbf{Are those numbers always powers of primes?}$. (The next term takes forever to calculate, but maybe I should wait longer before asking this question....)


      I dont know how complicated those questions are, so partial answers are also welcome.







      co.combinatorics gr.group-theory rt.representation-theory finite-groups






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      share|cite|improve this question













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      edited Nov 4 at 0:29

























      asked Nov 3 at 22:22









      Mare

      3,38821131




      3,38821131






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          13
          down vote













          For $K=mathbb C$, if $mathbb C Gcong prod_i M_{a_i}(mathbb C)$,
          then the sequence
          $d(G)=[a_1,ldots,a_m]$ is the sequence of character
          degrees of the irreducible representations of $G$ over $mathbb C$.
          That is, it is the sequence of numbers in the first column
          of the character table for $G$.




          There is a lot that is known and a lot that is
          unknown about these sequences. For example:







          1. $|G|=a_1^2+cdots+a_m^2$.




          2. $m$ is the number of conjugacy classes of $G$.





          3. $a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.







          4. $|G/G'|$ is the number of $a_i$ equal to $1$. In particular,
            at least one $a_i$ is $1$, and the number of $1$'s is a divisor of $a_1^2+cdots+a_m^2$.





          5. There exists nonisomorphic groups $G$ and $H$ with
            $d(G)=d(H)$. It is even possible
            to choose one to be solvable and the other to be nonsolvable.





          6. If $sum_{i=1}^m a_i <16m/5$
            and $p$ is any prime that divides $sum_{i=1}^m a_i^2$, then $mgeq 2sqrt{p-1}$.







          The McKay Conjecture can be phrased as a statement about how the sequence $d(G)=[a_1,ldots,a_m]$ is related to the corresponding sequences for normalizers of Sylows. For a prime $p$, let $i_{p'}(G)$ be the number of $a_i$ such that $p$ does not divide $a_i$. The McKay conjecture is that $i_{p'}(G)=i_{p'}(N)$ if $N$ is a normalizer of a Sylow $p$-subgroup of $G$.





          None of this answers your questions. I'm only saying that the questions are hard.






          share|cite|improve this answer























          • Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
            – Mare
            Nov 4 at 0:31


















          up vote
          7
          down vote













          In addition to Keith's answer I would like to mention a problem posed by Richard Brauer.



          In [Brauer, Richard. Representations of finite groups. 1963 Lectures on
          Modern Mathematics, Vol. I pp. 133–175 Wiley, New York 20.80] one finds the following question:




          What are the possible complex group algebras of finite groups?




          For example $mathbb{C}^5times M_5(mathbb{C})$ is not a complex group algebra. To prove this we list the degrees
          of irreducible characters of groups of order $30$. We see that there are four
          groups of order $30$ and none of them has a group algebra isomorphic to our algebra.



          gap> n := 30;; for G in AllGroups(Size, n) do Print(CharacterDegrees(G), "n"); od;


          The answer is



          [ [ 1, 10 ], [ 2, 5 ] ] [ [ 1, 6 ], [ 2, 6 ] ] [ [ 1, 2 ], [ 2, 7 ] ] [ [ 1, 30 ] ]


          and this shows that the groups algebras of groups of order $30$
          are
          $$mathbb{C}^{10}times M_2(mathbb{C})^5,quad
          mathbb{C}^{6}times M_2(mathbb{C})^6,quad
          mathbb{C}^{2}times M_2(mathbb{C})^7,quad
          mathbb{C}^{30}.$$



          Brauer's question seems to be a very hard to attack with the existing methods. An interesting paper in this direction is [Moretó, Alexander(E-VLNC-AG). Complex group algebras of finite groups: Brauer's problem 1. (English summary) Adv. Math. 208 (2007), no. 1, 236–248]. Here is the link to Moretó's paper.






          share|cite|improve this answer

















          • 1




            I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
            – Mark Wildon
            Nov 4 at 14:38




















          up vote
          6
          down vote













          Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.






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            3 Answers
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            active

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            3 Answers
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            active

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            active

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            up vote
            13
            down vote













            For $K=mathbb C$, if $mathbb C Gcong prod_i M_{a_i}(mathbb C)$,
            then the sequence
            $d(G)=[a_1,ldots,a_m]$ is the sequence of character
            degrees of the irreducible representations of $G$ over $mathbb C$.
            That is, it is the sequence of numbers in the first column
            of the character table for $G$.




            There is a lot that is known and a lot that is
            unknown about these sequences. For example:







            1. $|G|=a_1^2+cdots+a_m^2$.




            2. $m$ is the number of conjugacy classes of $G$.





            3. $a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.







            4. $|G/G'|$ is the number of $a_i$ equal to $1$. In particular,
              at least one $a_i$ is $1$, and the number of $1$'s is a divisor of $a_1^2+cdots+a_m^2$.





            5. There exists nonisomorphic groups $G$ and $H$ with
              $d(G)=d(H)$. It is even possible
              to choose one to be solvable and the other to be nonsolvable.





            6. If $sum_{i=1}^m a_i <16m/5$
              and $p$ is any prime that divides $sum_{i=1}^m a_i^2$, then $mgeq 2sqrt{p-1}$.







            The McKay Conjecture can be phrased as a statement about how the sequence $d(G)=[a_1,ldots,a_m]$ is related to the corresponding sequences for normalizers of Sylows. For a prime $p$, let $i_{p'}(G)$ be the number of $a_i$ such that $p$ does not divide $a_i$. The McKay conjecture is that $i_{p'}(G)=i_{p'}(N)$ if $N$ is a normalizer of a Sylow $p$-subgroup of $G$.





            None of this answers your questions. I'm only saying that the questions are hard.






            share|cite|improve this answer























            • Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
              – Mare
              Nov 4 at 0:31















            up vote
            13
            down vote













            For $K=mathbb C$, if $mathbb C Gcong prod_i M_{a_i}(mathbb C)$,
            then the sequence
            $d(G)=[a_1,ldots,a_m]$ is the sequence of character
            degrees of the irreducible representations of $G$ over $mathbb C$.
            That is, it is the sequence of numbers in the first column
            of the character table for $G$.




            There is a lot that is known and a lot that is
            unknown about these sequences. For example:







            1. $|G|=a_1^2+cdots+a_m^2$.




            2. $m$ is the number of conjugacy classes of $G$.





            3. $a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.







            4. $|G/G'|$ is the number of $a_i$ equal to $1$. In particular,
              at least one $a_i$ is $1$, and the number of $1$'s is a divisor of $a_1^2+cdots+a_m^2$.





            5. There exists nonisomorphic groups $G$ and $H$ with
              $d(G)=d(H)$. It is even possible
              to choose one to be solvable and the other to be nonsolvable.





            6. If $sum_{i=1}^m a_i <16m/5$
              and $p$ is any prime that divides $sum_{i=1}^m a_i^2$, then $mgeq 2sqrt{p-1}$.







            The McKay Conjecture can be phrased as a statement about how the sequence $d(G)=[a_1,ldots,a_m]$ is related to the corresponding sequences for normalizers of Sylows. For a prime $p$, let $i_{p'}(G)$ be the number of $a_i$ such that $p$ does not divide $a_i$. The McKay conjecture is that $i_{p'}(G)=i_{p'}(N)$ if $N$ is a normalizer of a Sylow $p$-subgroup of $G$.





            None of this answers your questions. I'm only saying that the questions are hard.






            share|cite|improve this answer























            • Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
              – Mare
              Nov 4 at 0:31













            up vote
            13
            down vote










            up vote
            13
            down vote









            For $K=mathbb C$, if $mathbb C Gcong prod_i M_{a_i}(mathbb C)$,
            then the sequence
            $d(G)=[a_1,ldots,a_m]$ is the sequence of character
            degrees of the irreducible representations of $G$ over $mathbb C$.
            That is, it is the sequence of numbers in the first column
            of the character table for $G$.




            There is a lot that is known and a lot that is
            unknown about these sequences. For example:







            1. $|G|=a_1^2+cdots+a_m^2$.




            2. $m$ is the number of conjugacy classes of $G$.





            3. $a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.







            4. $|G/G'|$ is the number of $a_i$ equal to $1$. In particular,
              at least one $a_i$ is $1$, and the number of $1$'s is a divisor of $a_1^2+cdots+a_m^2$.





            5. There exists nonisomorphic groups $G$ and $H$ with
              $d(G)=d(H)$. It is even possible
              to choose one to be solvable and the other to be nonsolvable.





            6. If $sum_{i=1}^m a_i <16m/5$
              and $p$ is any prime that divides $sum_{i=1}^m a_i^2$, then $mgeq 2sqrt{p-1}$.







            The McKay Conjecture can be phrased as a statement about how the sequence $d(G)=[a_1,ldots,a_m]$ is related to the corresponding sequences for normalizers of Sylows. For a prime $p$, let $i_{p'}(G)$ be the number of $a_i$ such that $p$ does not divide $a_i$. The McKay conjecture is that $i_{p'}(G)=i_{p'}(N)$ if $N$ is a normalizer of a Sylow $p$-subgroup of $G$.





            None of this answers your questions. I'm only saying that the questions are hard.






            share|cite|improve this answer














            For $K=mathbb C$, if $mathbb C Gcong prod_i M_{a_i}(mathbb C)$,
            then the sequence
            $d(G)=[a_1,ldots,a_m]$ is the sequence of character
            degrees of the irreducible representations of $G$ over $mathbb C$.
            That is, it is the sequence of numbers in the first column
            of the character table for $G$.




            There is a lot that is known and a lot that is
            unknown about these sequences. For example:







            1. $|G|=a_1^2+cdots+a_m^2$.




            2. $m$ is the number of conjugacy classes of $G$.





            3. $a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.







            4. $|G/G'|$ is the number of $a_i$ equal to $1$. In particular,
              at least one $a_i$ is $1$, and the number of $1$'s is a divisor of $a_1^2+cdots+a_m^2$.





            5. There exists nonisomorphic groups $G$ and $H$ with
              $d(G)=d(H)$. It is even possible
              to choose one to be solvable and the other to be nonsolvable.





            6. If $sum_{i=1}^m a_i <16m/5$
              and $p$ is any prime that divides $sum_{i=1}^m a_i^2$, then $mgeq 2sqrt{p-1}$.







            The McKay Conjecture can be phrased as a statement about how the sequence $d(G)=[a_1,ldots,a_m]$ is related to the corresponding sequences for normalizers of Sylows. For a prime $p$, let $i_{p'}(G)$ be the number of $a_i$ such that $p$ does not divide $a_i$. The McKay conjecture is that $i_{p'}(G)=i_{p'}(N)$ if $N$ is a normalizer of a Sylow $p$-subgroup of $G$.





            None of this answers your questions. I'm only saying that the questions are hard.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 3 at 23:22

























            answered Nov 3 at 22:59









            Keith Kearnes

            6,25222946




            6,25222946












            • Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
              – Mare
              Nov 4 at 0:31


















            • Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
              – Mare
              Nov 4 at 0:31
















            Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
            – Mare
            Nov 4 at 0:31




            Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
            – Mare
            Nov 4 at 0:31










            up vote
            7
            down vote













            In addition to Keith's answer I would like to mention a problem posed by Richard Brauer.



            In [Brauer, Richard. Representations of finite groups. 1963 Lectures on
            Modern Mathematics, Vol. I pp. 133–175 Wiley, New York 20.80] one finds the following question:




            What are the possible complex group algebras of finite groups?




            For example $mathbb{C}^5times M_5(mathbb{C})$ is not a complex group algebra. To prove this we list the degrees
            of irreducible characters of groups of order $30$. We see that there are four
            groups of order $30$ and none of them has a group algebra isomorphic to our algebra.



            gap> n := 30;; for G in AllGroups(Size, n) do Print(CharacterDegrees(G), "n"); od;


            The answer is



            [ [ 1, 10 ], [ 2, 5 ] ] [ [ 1, 6 ], [ 2, 6 ] ] [ [ 1, 2 ], [ 2, 7 ] ] [ [ 1, 30 ] ]


            and this shows that the groups algebras of groups of order $30$
            are
            $$mathbb{C}^{10}times M_2(mathbb{C})^5,quad
            mathbb{C}^{6}times M_2(mathbb{C})^6,quad
            mathbb{C}^{2}times M_2(mathbb{C})^7,quad
            mathbb{C}^{30}.$$



            Brauer's question seems to be a very hard to attack with the existing methods. An interesting paper in this direction is [Moretó, Alexander(E-VLNC-AG). Complex group algebras of finite groups: Brauer's problem 1. (English summary) Adv. Math. 208 (2007), no. 1, 236–248]. Here is the link to Moretó's paper.






            share|cite|improve this answer

















            • 1




              I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
              – Mark Wildon
              Nov 4 at 14:38

















            up vote
            7
            down vote













            In addition to Keith's answer I would like to mention a problem posed by Richard Brauer.



            In [Brauer, Richard. Representations of finite groups. 1963 Lectures on
            Modern Mathematics, Vol. I pp. 133–175 Wiley, New York 20.80] one finds the following question:




            What are the possible complex group algebras of finite groups?




            For example $mathbb{C}^5times M_5(mathbb{C})$ is not a complex group algebra. To prove this we list the degrees
            of irreducible characters of groups of order $30$. We see that there are four
            groups of order $30$ and none of them has a group algebra isomorphic to our algebra.



            gap> n := 30;; for G in AllGroups(Size, n) do Print(CharacterDegrees(G), "n"); od;


            The answer is



            [ [ 1, 10 ], [ 2, 5 ] ] [ [ 1, 6 ], [ 2, 6 ] ] [ [ 1, 2 ], [ 2, 7 ] ] [ [ 1, 30 ] ]


            and this shows that the groups algebras of groups of order $30$
            are
            $$mathbb{C}^{10}times M_2(mathbb{C})^5,quad
            mathbb{C}^{6}times M_2(mathbb{C})^6,quad
            mathbb{C}^{2}times M_2(mathbb{C})^7,quad
            mathbb{C}^{30}.$$



            Brauer's question seems to be a very hard to attack with the existing methods. An interesting paper in this direction is [Moretó, Alexander(E-VLNC-AG). Complex group algebras of finite groups: Brauer's problem 1. (English summary) Adv. Math. 208 (2007), no. 1, 236–248]. Here is the link to Moretó's paper.






            share|cite|improve this answer

















            • 1




              I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
              – Mark Wildon
              Nov 4 at 14:38















            up vote
            7
            down vote










            up vote
            7
            down vote









            In addition to Keith's answer I would like to mention a problem posed by Richard Brauer.



            In [Brauer, Richard. Representations of finite groups. 1963 Lectures on
            Modern Mathematics, Vol. I pp. 133–175 Wiley, New York 20.80] one finds the following question:




            What are the possible complex group algebras of finite groups?




            For example $mathbb{C}^5times M_5(mathbb{C})$ is not a complex group algebra. To prove this we list the degrees
            of irreducible characters of groups of order $30$. We see that there are four
            groups of order $30$ and none of them has a group algebra isomorphic to our algebra.



            gap> n := 30;; for G in AllGroups(Size, n) do Print(CharacterDegrees(G), "n"); od;


            The answer is



            [ [ 1, 10 ], [ 2, 5 ] ] [ [ 1, 6 ], [ 2, 6 ] ] [ [ 1, 2 ], [ 2, 7 ] ] [ [ 1, 30 ] ]


            and this shows that the groups algebras of groups of order $30$
            are
            $$mathbb{C}^{10}times M_2(mathbb{C})^5,quad
            mathbb{C}^{6}times M_2(mathbb{C})^6,quad
            mathbb{C}^{2}times M_2(mathbb{C})^7,quad
            mathbb{C}^{30}.$$



            Brauer's question seems to be a very hard to attack with the existing methods. An interesting paper in this direction is [Moretó, Alexander(E-VLNC-AG). Complex group algebras of finite groups: Brauer's problem 1. (English summary) Adv. Math. 208 (2007), no. 1, 236–248]. Here is the link to Moretó's paper.






            share|cite|improve this answer












            In addition to Keith's answer I would like to mention a problem posed by Richard Brauer.



            In [Brauer, Richard. Representations of finite groups. 1963 Lectures on
            Modern Mathematics, Vol. I pp. 133–175 Wiley, New York 20.80] one finds the following question:




            What are the possible complex group algebras of finite groups?




            For example $mathbb{C}^5times M_5(mathbb{C})$ is not a complex group algebra. To prove this we list the degrees
            of irreducible characters of groups of order $30$. We see that there are four
            groups of order $30$ and none of them has a group algebra isomorphic to our algebra.



            gap> n := 30;; for G in AllGroups(Size, n) do Print(CharacterDegrees(G), "n"); od;


            The answer is



            [ [ 1, 10 ], [ 2, 5 ] ] [ [ 1, 6 ], [ 2, 6 ] ] [ [ 1, 2 ], [ 2, 7 ] ] [ [ 1, 30 ] ]


            and this shows that the groups algebras of groups of order $30$
            are
            $$mathbb{C}^{10}times M_2(mathbb{C})^5,quad
            mathbb{C}^{6}times M_2(mathbb{C})^6,quad
            mathbb{C}^{2}times M_2(mathbb{C})^7,quad
            mathbb{C}^{30}.$$



            Brauer's question seems to be a very hard to attack with the existing methods. An interesting paper in this direction is [Moretó, Alexander(E-VLNC-AG). Complex group algebras of finite groups: Brauer's problem 1. (English summary) Adv. Math. 208 (2007), no. 1, 236–248]. Here is the link to Moretó's paper.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 4 at 12:41









            Leandro Vendramin

            2,44011727




            2,44011727








            • 1




              I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
              – Mark Wildon
              Nov 4 at 14:38
















            • 1




              I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
              – Mark Wildon
              Nov 4 at 14:38










            1




            1




            I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
            – Mark Wildon
            Nov 4 at 14:38






            I've expanded on your final paragraph in my answer (which was a bit too long for a comment).
            – Mark Wildon
            Nov 4 at 14:38












            up vote
            6
            down vote













            Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.






            share|cite|improve this answer

























              up vote
              6
              down vote













              Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.






              share|cite|improve this answer























                up vote
                6
                down vote










                up vote
                6
                down vote









                Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.






                share|cite|improve this answer












                Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 4 at 14:36









                Mark Wildon

                5,81212843




                5,81212843






























                     

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