Which partitions realise group algebras of finite groups?
up vote
12
down vote
favorite
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:
Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?
Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?
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
$textbf{Are those numbers always powers of primes?}$. (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
add a comment |
up vote
12
down vote
favorite
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:
Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?
Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?
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
$textbf{Are those numbers always powers of primes?}$. (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
add a comment |
up vote
12
down vote
favorite
up vote
12
down vote
favorite
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:
Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?
Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?
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
$textbf{Are those numbers always powers of primes?}$. (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
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:
Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$?
Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra?
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
$textbf{Are those numbers always powers of primes?}$. (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
co.combinatorics gr.group-theory rt.representation-theory finite-groups
edited Nov 4 at 0:29
asked Nov 3 at 22:22
Mare
3,38821131
3,38821131
add a comment |
add a comment |
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:
$|G|=a_1^2+cdots+a_m^2$.
- $m$ is the number of conjugacy classes of $G$.
$a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.
$|G/G'|$ 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 $a_1^2+cdots+a_m^2$.
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.
- 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.
Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
– Mare
Nov 4 at 0:31
add a comment |
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.
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
add a comment |
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.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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:
$|G|=a_1^2+cdots+a_m^2$.
- $m$ is the number of conjugacy classes of $G$.
$a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.
$|G/G'|$ 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 $a_1^2+cdots+a_m^2$.
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.
- 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.
Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
– Mare
Nov 4 at 0:31
add a comment |
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:
$|G|=a_1^2+cdots+a_m^2$.
- $m$ is the number of conjugacy classes of $G$.
$a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.
$|G/G'|$ 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 $a_1^2+cdots+a_m^2$.
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.
- 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.
Thanks for your answer. In particular, 6. looks new to me. I added some more specific questions.
– Mare
Nov 4 at 0:31
add a comment |
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:
$|G|=a_1^2+cdots+a_m^2$.
- $m$ is the number of conjugacy classes of $G$.
$a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.
$|G/G'|$ 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 $a_1^2+cdots+a_m^2$.
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.
- 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.
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:
$|G|=a_1^2+cdots+a_m^2$.
- $m$ is the number of conjugacy classes of $G$.
$a_i$ divides $a_1^2+cdots+a_m^2$ for each $i$.
$|G/G'|$ 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 $a_1^2+cdots+a_m^2$.
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.
- 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.
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
add a comment |
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
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 4 at 14:36
Mark Wildon
5,81212843
5,81212843
add a comment |
add a comment |
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