Group of order $pq$ (both primes) where $pnmid q-1,~qnmid p-1$
$begingroup$
This little number theory problem came from an application of Sylow theorem which said for groups of order $pq$, where $p,~q$ are primes and $pnmid q-1,~qnmid p-1$. Then $G$ is cyclic. By Sylow III Thm, $n_p=pk+1mid pq$, but why this implies $n_p=1$? I can't see the clear reason.
abstract-algebra group-theory number-theory
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
$endgroup$
add a comment |
$begingroup$
This little number theory problem came from an application of Sylow theorem which said for groups of order $pq$, where $p,~q$ are primes and $pnmid q-1,~qnmid p-1$. Then $G$ is cyclic. By Sylow III Thm, $n_p=pk+1mid pq$, but why this implies $n_p=1$? I can't see the clear reason.
abstract-algebra group-theory number-theory
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
$endgroup$
3
$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29
add a comment |
$begingroup$
This little number theory problem came from an application of Sylow theorem which said for groups of order $pq$, where $p,~q$ are primes and $pnmid q-1,~qnmid p-1$. Then $G$ is cyclic. By Sylow III Thm, $n_p=pk+1mid pq$, but why this implies $n_p=1$? I can't see the clear reason.
abstract-algebra group-theory number-theory
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
$endgroup$
This little number theory problem came from an application of Sylow theorem which said for groups of order $pq$, where $p,~q$ are primes and $pnmid q-1,~qnmid p-1$. Then $G$ is cyclic. By Sylow III Thm, $n_p=pk+1mid pq$, but why this implies $n_p=1$? I can't see the clear reason.
abstract-algebra group-theory number-theory
abstract-algebra group-theory number-theory
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
edited Dec 24 '18 at 13:12
Shaun
11.1k113688
11.1k113688
asked Dec 24 '18 at 4:26
EricEric
1,851615
asked Dec 24 '18 at 4:26
EricEric
1,851615
asked Dec 24 '18 at 4:26
EricEric
1,851615
1,851615
3
$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29
add a comment |
3
$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29
3
3
$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29
$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.
Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) in S_p times S_q mapsto st in G$ is an isomorphism. Consequently, $G simeq S_p times S_q simeq mathbb{Z}_p oplus mathbb{Z}_q stackrel{(CRT)}{simeq} mathbb{Z}_{pq}.$
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
$endgroup$
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
add a comment |
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$begingroup$
There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.
Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) in S_p times S_q mapsto st in G$ is an isomorphism. Consequently, $G simeq S_p times S_q simeq mathbb{Z}_p oplus mathbb{Z}_q stackrel{(CRT)}{simeq} mathbb{Z}_{pq}.$
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
$endgroup$
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
add a comment |
$begingroup$
There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.
Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) in S_p times S_q mapsto st in G$ is an isomorphism. Consequently, $G simeq S_p times S_q simeq mathbb{Z}_p oplus mathbb{Z}_q stackrel{(CRT)}{simeq} mathbb{Z}_{pq}.$
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
$endgroup$
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
add a comment |
$begingroup$
There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.
Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) in S_p times S_q mapsto st in G$ is an isomorphism. Consequently, $G simeq S_p times S_q simeq mathbb{Z}_p oplus mathbb{Z}_q stackrel{(CRT)}{simeq} mathbb{Z}_{pq}.$
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
$endgroup$
There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.
Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) in S_p times S_q mapsto st in G$ is an isomorphism. Consequently, $G simeq S_p times S_q simeq mathbb{Z}_p oplus mathbb{Z}_q stackrel{(CRT)}{simeq} mathbb{Z}_{pq}.$
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
edited Dec 24 '18 at 4:51
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 4:32
Guido A.Guido A.
8,3351730
8,3351730
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
add a comment |
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
$begingroup$
Thank you very much, very clear!
$endgroup$
– Eric
Dec 24 '18 at 6:23
add a comment |
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$begingroup$
Being $gcd(n_p,p)=1$ implies $n_p$ divides $q$
$endgroup$
– Chinnapparaj R
Dec 24 '18 at 4:29