For $A,B in mathscr{M}_{2times2}(mathbb{Q}) $ of finite order, show that $AB$ has infinite order [duplicate]
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This question already has an answer here:
Examples and further results about the order of the product of two elements in a group
6 answers
Let $G$ be the group $ ( mathscr{M}_{2times2}(mathbb{Q}) , times ) $ of nonsingular matrices.
Let $ A = left (
begin{matrix}
0 & -1 \
1 & 0
end{matrix}
right ) $, the order of $A$ is $4$;
Let $ B = left (
begin{matrix}
0 & 1 \
-1 & -1
end{matrix}
right ) $, the order of $B$ is $3$.
Show that $AB$ has infinite order.
The only reasoning possible here is by contradiction as $G$ is not abelian. And so I tried, but I got stuck before any concrete development.
Any hints are welcome,
Thanks.
matrices group-theory cyclic-groups
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
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marked as duplicate by Dietrich Burde group-theory
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Jan 2 at 19:53
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.
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show 4 more comments
$begingroup$
This question already has an answer here:
Examples and further results about the order of the product of two elements in a group
6 answers
Let $G$ be the group $ ( mathscr{M}_{2times2}(mathbb{Q}) , times ) $ of nonsingular matrices.
Let $ A = left (
begin{matrix}
0 & -1 \
1 & 0
end{matrix}
right ) $, the order of $A$ is $4$;
Let $ B = left (
begin{matrix}
0 & 1 \
-1 & -1
end{matrix}
right ) $, the order of $B$ is $3$.
Show that $AB$ has infinite order.
The only reasoning possible here is by contradiction as $G$ is not abelian. And so I tried, but I got stuck before any concrete development.
Any hints are welcome,
Thanks.
matrices group-theory cyclic-groups
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
$endgroup$
marked as duplicate by Dietrich Burde group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Jan 2 at 19:53
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
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
1
$begingroup$
@GitGudmathscris where it's at
$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
1
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01
|
show 4 more comments
$begingroup$
This question already has an answer here:
Examples and further results about the order of the product of two elements in a group
6 answers
Let $G$ be the group $ ( mathscr{M}_{2times2}(mathbb{Q}) , times ) $ of nonsingular matrices.
Let $ A = left (
begin{matrix}
0 & -1 \
1 & 0
end{matrix}
right ) $, the order of $A$ is $4$;
Let $ B = left (
begin{matrix}
0 & 1 \
-1 & -1
end{matrix}
right ) $, the order of $B$ is $3$.
Show that $AB$ has infinite order.
The only reasoning possible here is by contradiction as $G$ is not abelian. And so I tried, but I got stuck before any concrete development.
Any hints are welcome,
Thanks.
matrices group-theory cyclic-groups
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
$endgroup$
This question already has an answer here:
Examples and further results about the order of the product of two elements in a group
6 answers
Let $G$ be the group $ ( mathscr{M}_{2times2}(mathbb{Q}) , times ) $ of nonsingular matrices.
Let $ A = left (
begin{matrix}
0 & -1 \
1 & 0
end{matrix}
right ) $, the order of $A$ is $4$;
Let $ B = left (
begin{matrix}
0 & 1 \
-1 & -1
end{matrix}
right ) $, the order of $B$ is $3$.
Show that $AB$ has infinite order.
The only reasoning possible here is by contradiction as $G$ is not abelian. And so I tried, but I got stuck before any concrete development.
Any hints are welcome,
Thanks.
This question already has an answer here:
Examples and further results about the order of the product of two elements in a group
6 answers
matrices group-theory cyclic-groups
matrices group-theory cyclic-groups
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
asked Jan 2 at 15:30
freehumoristfreehumorist
351214
351214
marked as duplicate by Dietrich Burde group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Jan 2 at 19:53
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 Dietrich Burde group-theory
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Jan 2 at 19:53
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
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
1
$begingroup$
@GitGudmathscris where it's at
$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
1
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01
|
show 4 more comments
2
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
1
$begingroup$
@GitGudmathscris where it's at
$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
1
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01
2
2
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
1
1
$begingroup$
@GitGud
mathscr is where it's at$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud
mathscr is where it's at$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
1
1
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
We compute
$$
AB = pmatrix{1&1\0&1}
$$
We can prove (using induction, for instance) that
$$
(AB)^n = pmatrix{1&n\0&1}
$$
Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
add a comment |
$begingroup$
You have $AB=begin{pmatrix}1&1\0&1end{pmatrix}$ And $(AB)^n=begin{pmatrix}1&n\0&1end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We compute
$$
AB = pmatrix{1&1\0&1}
$$
We can prove (using induction, for instance) that
$$
(AB)^n = pmatrix{1&n\0&1}
$$
Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
add a comment |
$begingroup$
We compute
$$
AB = pmatrix{1&1\0&1}
$$
We can prove (using induction, for instance) that
$$
(AB)^n = pmatrix{1&n\0&1}
$$
Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
add a comment |
$begingroup$
We compute
$$
AB = pmatrix{1&1\0&1}
$$
We can prove (using induction, for instance) that
$$
(AB)^n = pmatrix{1&n\0&1}
$$
Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
We compute
$$
AB = pmatrix{1&1\0&1}
$$
We can prove (using induction, for instance) that
$$
(AB)^n = pmatrix{1&n\0&1}
$$
Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
answered Jan 2 at 15:33
OmnomnomnomOmnomnomnom
129k792186
129k792186
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
add a comment |
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
Same answer, faster by the way than mine.
$endgroup$
– vidyarthi
Jan 2 at 15:39
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
$begingroup$
@vidyarthi and how can we this fast conclude that $(AB)^n = left ( begin{matrix} 1 & n \ 0 & 1 end{matrix} right ) $?
$endgroup$
– freehumorist
Jan 2 at 15:56
1
1
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
$begingroup$
@freehumorist you just need to show that $$ pmatrix{1&1\0&1} pmatrix{1&n\0&1} = pmatrix{1&n+1\0&1}$$
$endgroup$
– Omnomnomnom
Jan 2 at 16:10
add a comment |
$begingroup$
You have $AB=begin{pmatrix}1&1\0&1end{pmatrix}$ And $(AB)^n=begin{pmatrix}1&n\0&1end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
$endgroup$
add a comment |
$begingroup$
You have $AB=begin{pmatrix}1&1\0&1end{pmatrix}$ And $(AB)^n=begin{pmatrix}1&n\0&1end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
$endgroup$
add a comment |
$begingroup$
You have $AB=begin{pmatrix}1&1\0&1end{pmatrix}$ And $(AB)^n=begin{pmatrix}1&n\0&1end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
$endgroup$
You have $AB=begin{pmatrix}1&1\0&1end{pmatrix}$ And $(AB)^n=begin{pmatrix}1&n\0&1end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
answered Jan 2 at 15:37
vidyarthividyarthi
3,0731833
3,0731833
add a comment |
add a comment |
2
$begingroup$
That M's got swag.
$endgroup$
– Git Gud
Jan 2 at 15:41
1
$begingroup$
@GitGud
mathscris where it's at$endgroup$
– Omnomnomnom
Jan 2 at 15:47
$begingroup$
@GitGud it is not for the sake of beauty, but specially used.
$endgroup$
– freehumorist
Jan 2 at 15:55
1
$begingroup$
The answer by amWhy (at the duplicate) contains the answers given below.
$endgroup$
– Dietrich Burde
Jan 2 at 19:54
$begingroup$
Also this post has the same matrices as above in the answer by mrs.
$endgroup$
– Dietrich Burde
Jan 2 at 20:01