Mixing
Prove that an abelian group of order $pq$, with $(p, q) = 1$ is cyclic if it contains elements of order $p$...
-2
$begingroup$
Let $G$ be an abelian group of order $pq$, with $gcd(p, q) = 1$.
Assume there exists
$a, b in G$ such that $|a| = p, |b| = q$. Prove that $G$ is cyclic.
abstract-algebra group-theory
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
$endgroup$
|
show 3 more comments
-2
$begingroup$
Let $G$ be an abelian group of order $pq$, with $gcd(p, q) = 1$.
Assume there exists
$a, b in G$ such that $|a| = p, |b| = q$. Prove that $G$ is cyclic.
abstract-algebra group-theory
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
$endgroup$
$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
3
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
1
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45
|
show 3 more comments
-2
-2
-2
1
$begingroup$
Let $G$ be an abelian group of order $pq$, with $gcd(p, q) = 1$.
Assume there exists
$a, b in G$ such that $|a| = p, |b| = q$. Prove that $G$ is cyclic.
abstract-algebra group-theory
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
$endgroup$
Let $G$ be an abelian group of order $pq$, with $gcd(p, q) = 1$.
Assume there exists
$a, b in G$ such that $|a| = p, |b| = q$. Prove that $G$ is cyclic.
abstract-algebra group-theory
abstract-algebra group-theory
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
edited Aug 8 '18 at 0:45
Alan Wang
4,8921133
4,8921133
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
asked Aug 7 '18 at 16:07
Juan David Cardona GutierrezJuan David Cardona Gutierrez
12
12
$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
3
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
1
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45
|
show 3 more comments
$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
3
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
1
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45
$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
3
3
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
1
1
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45
|
show 3 more comments
1 Answer
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0
$begingroup$
Let $H=langle a rangle $ and $K=langle b rangle$.
Note that $Hcap K=1$.
By product formula, $|HK|=|H||K|=pq=|G|$. Hence $G=HK$.
Since $H$ and $K$ are both normal subgroups of $G$, $Gcong Htimes Kcong Bbb{Z}_ptimes Bbb{Z}_q$.
Since $p$ and $q$ are relatively prime, $Gcong Bbb{Z}_{pq}$ and hence $G$ is cyclic.
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
$endgroup$
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
add a comment |
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0
$begingroup$
Let $H=langle a rangle $ and $K=langle b rangle$.
Note that $Hcap K=1$.
By product formula, $|HK|=|H||K|=pq=|G|$. Hence $G=HK$.
Since $H$ and $K$ are both normal subgroups of $G$, $Gcong Htimes Kcong Bbb{Z}_ptimes Bbb{Z}_q$.
Since $p$ and $q$ are relatively prime, $Gcong Bbb{Z}_{pq}$ and hence $G$ is cyclic.
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
$endgroup$
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
add a comment |
0
$begingroup$
Let $H=langle a rangle $ and $K=langle b rangle$.
Note that $Hcap K=1$.
By product formula, $|HK|=|H||K|=pq=|G|$. Hence $G=HK$.
Since $H$ and $K$ are both normal subgroups of $G$, $Gcong Htimes Kcong Bbb{Z}_ptimes Bbb{Z}_q$.
Since $p$ and $q$ are relatively prime, $Gcong Bbb{Z}_{pq}$ and hence $G$ is cyclic.
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
$endgroup$
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
add a comment |
0
0
0
$begingroup$
Let $H=langle a rangle $ and $K=langle b rangle$.
Note that $Hcap K=1$.
By product formula, $|HK|=|H||K|=pq=|G|$. Hence $G=HK$.
Since $H$ and $K$ are both normal subgroups of $G$, $Gcong Htimes Kcong Bbb{Z}_ptimes Bbb{Z}_q$.
Since $p$ and $q$ are relatively prime, $Gcong Bbb{Z}_{pq}$ and hence $G$ is cyclic.
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
$endgroup$
Let $H=langle a rangle $ and $K=langle b rangle$.
Note that $Hcap K=1$.
By product formula, $|HK|=|H||K|=pq=|G|$. Hence $G=HK$.
Since $H$ and $K$ are both normal subgroups of $G$, $Gcong Htimes Kcong Bbb{Z}_ptimes Bbb{Z}_q$.
Since $p$ and $q$ are relatively prime, $Gcong Bbb{Z}_{pq}$ and hence $G$ is cyclic.
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
edited Jan 27 at 2:26
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
answered Aug 7 '18 at 16:16
Alan WangAlan Wang
4,8921133
4,8921133
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
add a comment |
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
Yeah, your second suggestion is way easier, because it requires zero knowledge of internal direct product business and $G$ is abelian, so computing the order of $ab$ is a snap.
$endgroup$
– Randall
Aug 7 '18 at 18:35
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
How come that the divisors of $pq$ are only $1,p,q,pq$ when you only know that $gcd(p,q)$=1? Take for instance $p=8$ and $q=9$.
$endgroup$
– Traveler
Jan 27 at 2:10
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
$begingroup$
@Traveler Yeah you are right. I here assumed that $p$ and $q$ are both prime numbers.
$endgroup$
– Alan Wang
Jan 27 at 2:25
add a comment |
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$begingroup$
i need help, i do not know as start
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:08
3
$begingroup$
$ab?{}{}{}{}{}$
$endgroup$
– Lord Shark the Unknown
Aug 7 '18 at 16:08
$begingroup$
i do not understand you
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:10
$begingroup$
obviously is gcd( p, q) = 1
$endgroup$
– Juan David Cardona Gutierrez
Aug 7 '18 at 16:11
1
$begingroup$
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
$endgroup$
– Arnaud Mortier
Aug 7 '18 at 16:45