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?
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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
edited Apr 13 '17 at 12:20
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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
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.
add a comment |
$begingroup$
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
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
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Any nonabelian group of exponent 3, for example.
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– Najib Idrissi
Feb 16 '14 at 16:38
add a comment |
$begingroup$
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
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
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 16 '14 at 16:18
UntitledUntitled
183111
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 16 '14 at 16:18
UntitledUntitled
183111
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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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 $|G|=27$?
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
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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 $AB neq BA$.
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– Omnomnomnom
Feb 16 '14 at 16:50
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Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself.
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– Nicky Hekster
Feb 16 '14 at 17:04
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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.
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– 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
$begingroup$
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 $|G|=27$?
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
$begingroup$
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 |
$begingroup$
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 $|G|=27$?
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
$begingroup$
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 |
$begingroup$
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 $|G|=27$?
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
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 $|G|=27$?
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
answered Feb 16 '14 at 16:33
Nicky HeksterNicky Hekster
28.4k53456
28.4k53456
$begingroup$
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 |
2
$begingroup$
Any nonabelian group of exponent 3, for example.
$endgroup$
– Najib Idrissi
Feb 16 '14 at 16:38