Non-abelian group in which $forall_{a,bin G} (ab)^3=a^3b^3$ [duplicate]

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This question already has an answer here:

• 
  If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?
  
  1 answer
  
  

Give an example of a non-abelian group, in which $(ab)^3=a^3b^3$ for every element $a,b$ in $G$.

I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that $(ab)^3=a^3b^3 iff (ba)^2=a^2b^2$. But I can't come up with one single example.

group-theory abelian-groups

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edited Apr 13 '17 at 12:20

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asked Feb 16 '14 at 16:18

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marked as duplicate by Jack Schmidt, Hagen von Eitzen, user61527, Andrew D. Hwang, Davide Giraudo Feb 16 '14 at 18:12

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Any nonabelian group of exponent 3, for example.
  $endgroup$
  – Najib Idrissi
  Feb 16 '14 at 16:38
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?
  
  1 answer
  
  

Give an example of a non-abelian group, in which $(ab)^3=a^3b^3$ for every element $a,b$ in $G$.

I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that $(ab)^3=a^3b^3 iff (ba)^2=a^2b^2$. But I can't come up with one single example.

group-theory abelian-groups

share|cite|improve this question

edited Apr 13 '17 at 12:20

Community♦

1

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

$endgroup$

marked as duplicate by Jack Schmidt, Hagen von Eitzen, user61527, Andrew D. Hwang, Davide Giraudo Feb 16 '14 at 18:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Any nonabelian group of exponent 3, for example.
  $endgroup$
  – Najib Idrissi
  Feb 16 '14 at 16:38
  

  
  
  
  

add a comment | 

1

1

1

0

$begingroup$

This question already has an answer here:

• 
  If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?
  
  1 answer
  
  

Give an example of a non-abelian group, in which $(ab)^3=a^3b^3$ for every element $a,b$ in $G$.

I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that $(ab)^3=a^3b^3 iff (ba)^2=a^2b^2$. But I can't come up with one single example.

group-theory abelian-groups

share|cite|improve this question

edited Apr 13 '17 at 12:20

Community♦

1

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

$endgroup$

This question already has an answer here:

• 
  If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?
  
  1 answer
  
  

Give an example of a non-abelian group, in which $(ab)^3=a^3b^3$ for every element $a,b$ in $G$.

I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that $(ab)^3=a^3b^3 iff (ba)^2=a^2b^2$. But I can't come up with one single example.

This question already has an answer here:

• 
  If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?
  
  1 answer
  
  

group-theory abelian-groups

group-theory abelian-groups

share|cite|improve this question

edited Apr 13 '17 at 12:20

Community♦

1

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

share|cite|improve this question

edited Apr 13 '17 at 12:20

Community♦

1

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

share|cite|improve this question

share|cite|improve this question

edited Apr 13 '17 at 12:20

Community♦

1

edited Apr 13 '17 at 12:20

Community♦

1

edited Apr 13 '17 at 12:20

Community♦

1

1

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

asked Feb 16 '14 at 16:18

UntitledUntitled

183111

183111

marked as duplicate by Jack Schmidt, Hagen von Eitzen, user61527, Andrew D. Hwang, Davide Giraudo Feb 16 '14 at 18:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Jack Schmidt, Hagen von Eitzen, user61527, Andrew D. Hwang, Davide Giraudo Feb 16 '14 at 18:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Any nonabelian group of exponent 3, for example.
  $endgroup$
  – Najib Idrissi
  Feb 16 '14 at 16:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Any nonabelian group of exponent 3, for example.
  $endgroup$
  – Najib Idrissi
  Feb 16 '14 at 16:38
  

  
  
  
  

2

2

$begingroup$
Any nonabelian group of exponent 3, for example.
$endgroup$
– Najib Idrissi
Feb 16 '14 at 16:38

$begingroup$
Any nonabelian group of exponent 3, for example.
$endgroup$
– Najib Idrissi
Feb 16 '14 at 16:38

add a comment | 

1 Answer
1

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1

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Hint: try the group of triangular $3 times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

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answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 16:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
  $endgroup$
  – Nicky Hekster
  Feb 16 '14 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 17:11
  
  
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Hint: try the group of triangular $3 times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

share|cite|improve this answer

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 16:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
  $endgroup$
  – Nicky Hekster
  Feb 16 '14 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 17:11
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: try the group of triangular $3 times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

share|cite|improve this answer

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 16:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
  $endgroup$
  – Nicky Hekster
  Feb 16 '14 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 17:11
  
  
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Hint: try the group of triangular $3 times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

share|cite|improve this answer

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

$endgroup$

Hint: try the group of triangular $3 times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

share|cite|improve this answer

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

share|cite|improve this answer

share|cite|improve this answer

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

answered Feb 16 '14 at 16:33

Nicky HeksterNicky Hekster

28.5k63456

28.5k63456

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 16:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
  $endgroup$
  – Nicky Hekster
  Feb 16 '14 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 17:11
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 16:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
  $endgroup$
  – Nicky Hekster
  Feb 16 '14 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
  $endgroup$
  – Omnomnomnom
  Feb 16 '14 at 17:11
  
  
  

  
  
  
  

$begingroup$
I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
$endgroup$
– Omnomnomnom
Feb 16 '14 at 16:50

$begingroup$
I'd just like to add: as evidence that $G$ is not abelian, note that $A = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 \}$ and $B = pmatrix{ 1 & 1 & 0 \ 0 & 1 & 2 \ 0 & 0 & 1 \}$ satisfy $AB neq BA$.
$endgroup$
– Omnomnomnom
Feb 16 '14 at 16:50

$begingroup$
Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
$endgroup$
– Nicky Hekster
Feb 16 '14 at 17:04

$begingroup$
Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
$endgroup$
– Nicky Hekster
Feb 16 '14 at 17:04

$begingroup$
Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
$endgroup$
– Omnomnomnom
Feb 16 '14 at 17:11

$begingroup$
Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for.
$endgroup$
– Omnomnomnom
Feb 16 '14 at 17:11

add a comment | 

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