a and b commutes of order p q with p and q co-prime show that ab is order pq [duplicate]
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This question already has an answer here:
Show that the order of $atimes b$ is equal to $nm$ if gcd(n,m)=1 [duplicate]
1 answer
Proove if $operatorname{ord}(u)=r$ and $operatorname{ord}(v)=s$ then $operatorname{ord}(uv)=rs$ - my attempt
3 answers
Let $(G,*)$ be an abelian group with two elements a and b of order p,q co-prime.
What is the order of a*b ?
At most pq as $(a*b)^{pq}=a^{pq}*b^{pq}=1_G$
I guess it is exactly pq but I do not see why ?
abstract-algebra abelian-groups
asked Jan 8 '17 at 19:36
JoachimJoachim
588
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marked as duplicate by Dietrich Burde, Bill Dubuque abstract-algebra
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Jan 8 '17 at 19:59
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:
Show that the order of $atimes b$ is equal to $nm$ if gcd(n,m)=1 [duplicate]
1 answer
Proove if $operatorname{ord}(u)=r$ and $operatorname{ord}(v)=s$ then $operatorname{ord}(uv)=rs$ - my attempt
3 answers
Let $(G,*)$ be an abelian group with two elements a and b of order p,q co-prime.
What is the order of a*b ?
At most pq as $(a*b)^{pq}=a^{pq}*b^{pq}=1_G$
I guess it is exactly pq but I do not see why ?
abstract-algebra abelian-groups
asked Jan 8 '17 at 19:36
JoachimJoachim
588
$endgroup$
marked as duplicate by Dietrich Burde, Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Jan 8 '17 at 19:59
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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True but the answer gives only a hint that did not help me
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– Joachim
Jan 8 '17 at 19:55
1
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Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58
add a comment |
$begingroup$
This question already has an answer here:
Show that the order of $atimes b$ is equal to $nm$ if gcd(n,m)=1 [duplicate]
1 answer
Proove if $operatorname{ord}(u)=r$ and $operatorname{ord}(v)=s$ then $operatorname{ord}(uv)=rs$ - my attempt
3 answers
Let $(G,*)$ be an abelian group with two elements a and b of order p,q co-prime.
What is the order of a*b ?
At most pq as $(a*b)^{pq}=a^{pq}*b^{pq}=1_G$
I guess it is exactly pq but I do not see why ?
abstract-algebra abelian-groups
asked Jan 8 '17 at 19:36
JoachimJoachim
588
$endgroup$
This question already has an answer here:
Show that the order of $atimes b$ is equal to $nm$ if gcd(n,m)=1 [duplicate]
1 answer
Proove if $operatorname{ord}(u)=r$ and $operatorname{ord}(v)=s$ then $operatorname{ord}(uv)=rs$ - my attempt
3 answers
Let $(G,*)$ be an abelian group with two elements a and b of order p,q co-prime.
What is the order of a*b ?
At most pq as $(a*b)^{pq}=a^{pq}*b^{pq}=1_G$
I guess it is exactly pq but I do not see why ?
This question already has an answer here:
Show that the order of $atimes b$ is equal to $nm$ if gcd(n,m)=1 [duplicate]
1 answer
Proove if $operatorname{ord}(u)=r$ and $operatorname{ord}(v)=s$ then $operatorname{ord}(uv)=rs$ - my attempt
3 answers
abstract-algebra abelian-groups
abstract-algebra abelian-groups
asked Jan 8 '17 at 19:36
JoachimJoachim
588
asked Jan 8 '17 at 19:36
JoachimJoachim
588
asked Jan 8 '17 at 19:36
JoachimJoachim
588
asked Jan 8 '17 at 19:36
JoachimJoachim
588
asked Jan 8 '17 at 19:36
JoachimJoachim
588
588
marked as duplicate by Dietrich Burde, Bill Dubuque abstract-algebra
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Jan 8 '17 at 19:59
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, Bill Dubuque abstract-algebra
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Jan 8 '17 at 19:59
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.
$begingroup$
True but the answer gives only a hint that did not help me
$endgroup$
– Joachim
Jan 8 '17 at 19:55
1
$begingroup$
Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58
add a comment |
$begingroup$
True but the answer gives only a hint that did not help me
$endgroup$
– Joachim
Jan 8 '17 at 19:55
1
$begingroup$
Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58
$begingroup$
True but the answer gives only a hint that did not help me
$endgroup$
– Joachim
Jan 8 '17 at 19:55
$begingroup$
True but the answer gives only a hint that did not help me
$endgroup$
– Joachim
Jan 8 '17 at 19:55
1
1
$begingroup$
Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58
$begingroup$
Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $langle b rangle $ and $langle a rangle$. The intersection of these two subgroups is ${e}$ by Lagrange's theorem.
So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
$endgroup$
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $langle b rangle $ and $langle a rangle$. The intersection of these two subgroups is ${e}$ by Lagrange's theorem.
So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
$endgroup$
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
|
show 1 more comment
$begingroup$
notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $langle b rangle $ and $langle a rangle$. The intersection of these two subgroups is ${e}$ by Lagrange's theorem.
So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
$endgroup$
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
|
show 1 more comment
$begingroup$
notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $langle b rangle $ and $langle a rangle$. The intersection of these two subgroups is ${e}$ by Lagrange's theorem.
So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
$endgroup$
notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $langle b rangle $ and $langle a rangle$. The intersection of these two subgroups is ${e}$ by Lagrange's theorem.
So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
answered Jan 8 '17 at 19:40
Jorge FernándezJorge Fernández
75.1k1190191
75.1k1190191
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
|
show 1 more comment
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
why Lagrange imply the intersection to be {e} ?
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
p/#G and q/#G with p q coprime
$endgroup$
– Joachim
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
because the order of the subgroup must divide both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:46
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
$begingroup$
I got it thanks
$endgroup$
– Joachim
Jan 8 '17 at 19:47
1
1
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
$begingroup$
No, use the fact that $langle a rangle cap langle b rangle $ is a subgroup of both $langle a rangle $ and $langle b rangle$. Using Lagrang'es theorem twice tells you that $|langle a rangle cap langle brangle |$ is a divisor of both $p$ and $q$.
$endgroup$
– Jorge Fernández
Jan 8 '17 at 19:48
|
show 1 more comment
$begingroup$
True but the answer gives only a hint that did not help me
$endgroup$
– Joachim
Jan 8 '17 at 19:55
1
$begingroup$
Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too).
$endgroup$
– Dietrich Burde
Jan 8 '17 at 19:58