Prove that an abelian group of order $pq$, with $(p, q) = 1$ is cyclic if it contains elements of order $p$ and $q$.
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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 at 0:45
Alan Wang
4,436932
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
 |Â
show 3 more comments
up vote
-2
down vote
favorite
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 at 0:45
Alan Wang
4,436932
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
3
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
1
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45
 |Â
show 3 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
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 at 0:45
Alan Wang
4,436932
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
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 at 0:45
Alan Wang
4,436932
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
edited Aug 8 at 0:45
Alan Wang
4,436932
edited Aug 8 at 0:45
Alan Wang
4,436932
edited Aug 8 at 0:45
Alan Wang
4,436932
4,436932
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
asked Aug 7 at 16:07
Juan David Cardona Gutierrez
12
12
i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
3
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
1
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45
 |Â
show 3 more comments
i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
3
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
1
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45
i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
3
3
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
1
1
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
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 $G$, $Gcong Htimes Kcong BbbZ_ptimes BbbZ_q$.
Since $p$ and $q$ are relatively prime, $Gcong BbbZ_pq$ and hence $G$ is cyclic.
Of course there are many other ways to solve the problem. For example, you may try to show that the order of $ab$ is $pq$ and hence $G=langle abrangle$ is cyclic.
The second way is shown below:
Note that $(ab)^pq=a^pqb^pq=1cdot1=1$.
Hence $|ab|$ divides pq.
The possibilities are $1,p,q$ and $pq$.
If $|ab|=1$, then $ab=1$ which implies $a=b^-1$ and $p=|a|=|b^-1|=|b|=q$, a contradiction.
If $|ab|=p$, then $1=(ab)^p=a^pb^p=b^p$ which implies $q$ divides $p$, a contradiction.
Similarly, $|ab|=q$ would lead to a contradiction.
So the only possibility is $|ab|=pq$.
answered Aug 7 at 16:16
Alan Wang
4,436932
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.
– Randall
Aug 7 at 18:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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 $G$, $Gcong Htimes Kcong BbbZ_ptimes BbbZ_q$.
Since $p$ and $q$ are relatively prime, $Gcong BbbZ_pq$ and hence $G$ is cyclic.
Of course there are many other ways to solve the problem. For example, you may try to show that the order of $ab$ is $pq$ and hence $G=langle abrangle$ is cyclic.
The second way is shown below:
Note that $(ab)^pq=a^pqb^pq=1cdot1=1$.
Hence $|ab|$ divides pq.
The possibilities are $1,p,q$ and $pq$.
If $|ab|=1$, then $ab=1$ which implies $a=b^-1$ and $p=|a|=|b^-1|=|b|=q$, a contradiction.
If $|ab|=p$, then $1=(ab)^p=a^pb^p=b^p$ which implies $q$ divides $p$, a contradiction.
Similarly, $|ab|=q$ would lead to a contradiction.
So the only possibility is $|ab|=pq$.
answered Aug 7 at 16:16
Alan Wang
4,436932
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.
– Randall
Aug 7 at 18:35
add a comment |Â
up vote
0
down vote
accepted
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 $G$, $Gcong Htimes Kcong BbbZ_ptimes BbbZ_q$.
Since $p$ and $q$ are relatively prime, $Gcong BbbZ_pq$ and hence $G$ is cyclic.
Of course there are many other ways to solve the problem. For example, you may try to show that the order of $ab$ is $pq$ and hence $G=langle abrangle$ is cyclic.
The second way is shown below:
Note that $(ab)^pq=a^pqb^pq=1cdot1=1$.
Hence $|ab|$ divides pq.
The possibilities are $1,p,q$ and $pq$.
If $|ab|=1$, then $ab=1$ which implies $a=b^-1$ and $p=|a|=|b^-1|=|b|=q$, a contradiction.
If $|ab|=p$, then $1=(ab)^p=a^pb^p=b^p$ which implies $q$ divides $p$, a contradiction.
Similarly, $|ab|=q$ would lead to a contradiction.
So the only possibility is $|ab|=pq$.
answered Aug 7 at 16:16
Alan Wang
4,436932
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.
– Randall
Aug 7 at 18:35
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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 $G$, $Gcong Htimes Kcong BbbZ_ptimes BbbZ_q$.
Since $p$ and $q$ are relatively prime, $Gcong BbbZ_pq$ and hence $G$ is cyclic.
Of course there are many other ways to solve the problem. For example, you may try to show that the order of $ab$ is $pq$ and hence $G=langle abrangle$ is cyclic.
The second way is shown below:
Note that $(ab)^pq=a^pqb^pq=1cdot1=1$.
Hence $|ab|$ divides pq.
The possibilities are $1,p,q$ and $pq$.
If $|ab|=1$, then $ab=1$ which implies $a=b^-1$ and $p=|a|=|b^-1|=|b|=q$, a contradiction.
If $|ab|=p$, then $1=(ab)^p=a^pb^p=b^p$ which implies $q$ divides $p$, a contradiction.
Similarly, $|ab|=q$ would lead to a contradiction.
So the only possibility is $|ab|=pq$.
answered Aug 7 at 16:16
Alan Wang
4,436932
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 $G$, $Gcong Htimes Kcong BbbZ_ptimes BbbZ_q$.
Since $p$ and $q$ are relatively prime, $Gcong BbbZ_pq$ and hence $G$ is cyclic.
Of course there are many other ways to solve the problem. For example, you may try to show that the order of $ab$ is $pq$ and hence $G=langle abrangle$ is cyclic.
The second way is shown below:
Note that $(ab)^pq=a^pqb^pq=1cdot1=1$.
Hence $|ab|$ divides pq.
The possibilities are $1,p,q$ and $pq$.
If $|ab|=1$, then $ab=1$ which implies $a=b^-1$ and $p=|a|=|b^-1|=|b|=q$, a contradiction.
If $|ab|=p$, then $1=(ab)^p=a^pb^p=b^p$ which implies $q$ divides $p$, a contradiction.
Similarly, $|ab|=q$ would lead to a contradiction.
So the only possibility is $|ab|=pq$.
answered Aug 7 at 16:16
Alan Wang
4,436932
edited Aug 8 at 0:54
answered Aug 7 at 16:16
Alan Wang
4,436932
answered Aug 7 at 16:16
Alan Wang
4,436932
answered Aug 7 at 16:16
Alan Wang
4,436932
4,436932
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.
– Randall
Aug 7 at 18:35
add a comment |Â
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.
– Randall
Aug 7 at 18:35
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.
– Randall
Aug 7 at 18:35
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.
– Randall
Aug 7 at 18:35
add a comment |Â
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i need help, i do not know as start
– Juan David Cardona Gutierrez
Aug 7 at 16:08
3
$ab?$
– Lord Shark the Unknown
Aug 7 at 16:08
i do not understand you
– Juan David Cardona Gutierrez
Aug 7 at 16:10
obviously is gcd( p, q) = 1
– Juan David Cardona Gutierrez
Aug 7 at 16:11
1
@Batominovski you can get the answer by following this recipe: math.meta.stackexchange.com/questions/28795/…
– Arnaud Mortier
Aug 7 at 16:45