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Efficient computation of conjugacy classes of a small group.

Character Table From PresentationComputing Conjugacy Classes of Subgroups in GAPconjugacy class of a dicyclic groupConjugacy Classes and Character Degrees of E8(2)Character Table Dihedral group of $D_6$Character table for $G:= langle x,y mid x^5=y^4=1text and yx=x^2yrangle$Size of Conjugacy Classes group of order 168Rational Classes (Conjugacy Classes)Find conjugacy classes of $S_3$Character Table from Generators of a Group

5

1

$begingroup$

I'm trying to construct a character table for a group of order 54 given by:

$$ langle a,b : a^9 = b^6 = 1, b^-1 a b = a^2rangle $$

To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on how to compute those efficiently.

group-theory group-presentation combinatorial-group-theory

share|cite|improve this question

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There's always GAP.
  $endgroup$
  – Shaun
  May 24 at 13:13
  

  
  
  
  

add a comment | 

5

1

$begingroup$

I'm trying to construct a character table for a group of order 54 given by:

$$ langle a,b : a^9 = b^6 = 1, b^-1 a b = a^2rangle $$

To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on how to compute those efficiently.

group-theory group-presentation combinatorial-group-theory

share|cite|improve this question

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There's always GAP.
  $endgroup$
  – Shaun
  May 24 at 13:13
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to construct a character table for a group of order 54 given by:

$$ langle a,b : a^9 = b^6 = 1, b^-1 a b = a^2rangle $$

To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on how to compute those efficiently.

group-theory group-presentation combinatorial-group-theory

share|cite|improve this question

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

$endgroup$

share|cite|improve this question

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

share|cite|improve this question

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

edited May 24 at 13:12

Shaun

11.3k123688

edited May 24 at 13:12

Shaun

11.3k123688

asked May 24 at 10:34

RadostRadost

1,30416

asked May 24 at 10:34

RadostRadost

1,30416

2 Answers
2

active

oldest

votes

5

$begingroup$

$G = a^mb^n:0le mle 8,0le nle 5$. For an element $a^mb^nin G$, $b^-1a^mb^nb = a^2mb^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^-1a^mb^na = a^m-1b^nab^-nb^n = a^m-1a^5^nb^n$$since $ba^2b^-1 = a = (a^2)^5$ and $a^2$ is also a generator of $langle arangle$. Note that a conjugate by $a$ or $b$ does not change $n$. Therefore, we only need to discuss $n$.

(1) $n=0$, then $a^-1a^mb^na = a^mb^n$, so the conjugate by $a$ is fixed. So $1,a,a^2,a^4,a^5,a^7,a^8,a^3,a^6$ are conjugacy classes.

(2) $n=1$, then $a^-1a^mb^na = a^m+4b^n$. So $b,ab,dots, a^8b$ is a conjugacy class, since $m+4$ runs all the $m$'s.

(3) $n=2$, then $a^-1a^mb^na = a^m+24b^n = a^m+6b^n$. Now $a^3b^2$, $a^6b^2$ and $b^2$ are in the same conjugacy class, while other $m$ forms another conjugacy class.

(4) $n=3$, then $a^-1a^mb^na = a^m+124b^n = a^m+7b^n$. The same as (2) since it runs all the $m$'s.

(5) $n=4$, then $a^-1a^mb^na = a^m+3b^n $. The same as (3).

(6) $n=5$, then $a^-1a^mb^na = a^m+1b^n$. The same as (2).

share|cite|improve this answer

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe there are some mistakes in calculation, but the key is that the conjugate does not change $n$.
  $endgroup$
  – Hongyi Huang
  May 24 at 11:14
  

  
  
  
  

add a comment | 

5

$begingroup$

From the defining relations try to get a description of all the elements.

Any element is a "word" involving $a$ and $b$.

Last condition $b^-1ab = a^2$ can be rewritten as $ab= ba^2$. This modified version says whenever a comes ahead of b it can be pushed to come behind b, but as $a^2$. So all words in a and b can be reformulated to have b's in the front and a's at end. As $a^9=b^6=1$, any element can be written as $b^m a^n$ with $m=0,1,2,ldots, 8, n=0,1,2,ldots, 5$, giving 54 possibilities.

Now try to describe conjuates of elements of type $a^k$, then $b^k$ and then $a^k b^l$ by any $g$ ($g$ has to be of the form $b^m a^n$ as above). It is doable.

share|cite|improve this answer

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

$endgroup$

add a comment | 

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5

$begingroup$

$G = a^mb^n:0le mle 8,0le nle 5$. For an element $a^mb^nin G$, $b^-1a^mb^nb = a^2mb^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^-1a^mb^na = a^m-1b^nab^-nb^n = a^m-1a^5^nb^n$$since $ba^2b^-1 = a = (a^2)^5$ and $a^2$ is also a generator of $langle arangle$. Note that a conjugate by $a$ or $b$ does not change $n$. Therefore, we only need to discuss $n$.

(1) $n=0$, then $a^-1a^mb^na = a^mb^n$, so the conjugate by $a$ is fixed. So $1,a,a^2,a^4,a^5,a^7,a^8,a^3,a^6$ are conjugacy classes.

(2) $n=1$, then $a^-1a^mb^na = a^m+4b^n$. So $b,ab,dots, a^8b$ is a conjugacy class, since $m+4$ runs all the $m$'s.

(3) $n=2$, then $a^-1a^mb^na = a^m+24b^n = a^m+6b^n$. Now $a^3b^2$, $a^6b^2$ and $b^2$ are in the same conjugacy class, while other $m$ forms another conjugacy class.

(4) $n=3$, then $a^-1a^mb^na = a^m+124b^n = a^m+7b^n$. The same as (2) since it runs all the $m$'s.

(5) $n=4$, then $a^-1a^mb^na = a^m+3b^n $. The same as (3).

(6) $n=5$, then $a^-1a^mb^na = a^m+1b^n$. The same as (2).

share|cite|improve this answer

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe there are some mistakes in calculation, but the key is that the conjugate does not change $n$.
  $endgroup$
  – Hongyi Huang
  May 24 at 11:14
  

  
  
  
  

add a comment | 

5

$begingroup$

$G = a^mb^n:0le mle 8,0le nle 5$. For an element $a^mb^nin G$, $b^-1a^mb^nb = a^2mb^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^-1a^mb^na = a^m-1b^nab^-nb^n = a^m-1a^5^nb^n$$since $ba^2b^-1 = a = (a^2)^5$ and $a^2$ is also a generator of $langle arangle$. Note that a conjugate by $a$ or $b$ does not change $n$. Therefore, we only need to discuss $n$.

(1) $n=0$, then $a^-1a^mb^na = a^mb^n$, so the conjugate by $a$ is fixed. So $1,a,a^2,a^4,a^5,a^7,a^8,a^3,a^6$ are conjugacy classes.

(2) $n=1$, then $a^-1a^mb^na = a^m+4b^n$. So $b,ab,dots, a^8b$ is a conjugacy class, since $m+4$ runs all the $m$'s.

(3) $n=2$, then $a^-1a^mb^na = a^m+24b^n = a^m+6b^n$. Now $a^3b^2$, $a^6b^2$ and $b^2$ are in the same conjugacy class, while other $m$ forms another conjugacy class.

(4) $n=3$, then $a^-1a^mb^na = a^m+124b^n = a^m+7b^n$. The same as (2) since it runs all the $m$'s.

(5) $n=4$, then $a^-1a^mb^na = a^m+3b^n $. The same as (3).

(6) $n=5$, then $a^-1a^mb^na = a^m+1b^n$. The same as (2).

share|cite|improve this answer

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe there are some mistakes in calculation, but the key is that the conjugate does not change $n$.
  $endgroup$
  – Hongyi Huang
  May 24 at 11:14
  

  
  
  
  

add a comment | 

$begingroup$

$G = a^mb^n:0le mle 8,0le nle 5$. For an element $a^mb^nin G$, $b^-1a^mb^nb = a^2mb^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^-1a^mb^na = a^m-1b^nab^-nb^n = a^m-1a^5^nb^n$$since $ba^2b^-1 = a = (a^2)^5$ and $a^2$ is also a generator of $langle arangle$. Note that a conjugate by $a$ or $b$ does not change $n$. Therefore, we only need to discuss $n$.

(1) $n=0$, then $a^-1a^mb^na = a^mb^n$, so the conjugate by $a$ is fixed. So $1,a,a^2,a^4,a^5,a^7,a^8,a^3,a^6$ are conjugacy classes.

(2) $n=1$, then $a^-1a^mb^na = a^m+4b^n$. So $b,ab,dots, a^8b$ is a conjugacy class, since $m+4$ runs all the $m$'s.

(3) $n=2$, then $a^-1a^mb^na = a^m+24b^n = a^m+6b^n$. Now $a^3b^2$, $a^6b^2$ and $b^2$ are in the same conjugacy class, while other $m$ forms another conjugacy class.

(4) $n=3$, then $a^-1a^mb^na = a^m+124b^n = a^m+7b^n$. The same as (2) since it runs all the $m$'s.

(5) $n=4$, then $a^-1a^mb^na = a^m+3b^n $. The same as (3).

(6) $n=5$, then $a^-1a^mb^na = a^m+1b^n$. The same as (2).

share|cite|improve this answer

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

$endgroup$

share|cite|improve this answer

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

answered May 24 at 11:09

Hongyi HuangHongyi Huang

1,814116

5

$begingroup$

From the defining relations try to get a description of all the elements.

Any element is a "word" involving $a$ and $b$.

Last condition $b^-1ab = a^2$ can be rewritten as $ab= ba^2$. This modified version says whenever a comes ahead of b it can be pushed to come behind b, but as $a^2$. So all words in a and b can be reformulated to have b's in the front and a's at end. As $a^9=b^6=1$, any element can be written as $b^m a^n$ with $m=0,1,2,ldots, 8, n=0,1,2,ldots, 5$, giving 54 possibilities.

Now try to describe conjuates of elements of type $a^k$, then $b^k$ and then $a^k b^l$ by any $g$ ($g$ has to be of the form $b^m a^n$ as above). It is doable.

share|cite|improve this answer

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

$endgroup$

add a comment | 

5

$begingroup$

From the defining relations try to get a description of all the elements.

Any element is a "word" involving $a$ and $b$.

Last condition $b^-1ab = a^2$ can be rewritten as $ab= ba^2$. This modified version says whenever a comes ahead of b it can be pushed to come behind b, but as $a^2$. So all words in a and b can be reformulated to have b's in the front and a's at end. As $a^9=b^6=1$, any element can be written as $b^m a^n$ with $m=0,1,2,ldots, 8, n=0,1,2,ldots, 5$, giving 54 possibilities.

Now try to describe conjuates of elements of type $a^k$, then $b^k$ and then $a^k b^l$ by any $g$ ($g$ has to be of the form $b^m a^n$ as above). It is doable.

share|cite|improve this answer

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

$endgroup$

add a comment | 

$begingroup$

From the defining relations try to get a description of all the elements.

Any element is a "word" involving $a$ and $b$.

Last condition $b^-1ab = a^2$ can be rewritten as $ab= ba^2$. This modified version says whenever a comes ahead of b it can be pushed to come behind b, but as $a^2$. So all words in a and b can be reformulated to have b's in the front and a's at end. As $a^9=b^6=1$, any element can be written as $b^m a^n$ with $m=0,1,2,ldots, 8, n=0,1,2,ldots, 5$, giving 54 possibilities.

Now try to describe conjuates of elements of type $a^k$, then $b^k$ and then $a^k b^l$ by any $g$ ($g$ has to be of the form $b^m a^n$ as above). It is doable.

share|cite|improve this answer

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

$endgroup$

share|cite|improve this answer

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237

answered May 24 at 11:06

P VanchinathanP Vanchinathan

16k12237