Cardinality of Sylow Sub-Groups of a Group that has Cardinality of Prime Product

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Let $G$ be a group with $#G=p cdot q$ for $p,q in mathbbP$.



W.l.o.g. $p>q$. According to Sylow theorems we have one Sylow p-subgroup, let's call it $P$. Let $Q$ be a Sylow q-subgroup.



How can I prove that $#P=p$ and $#Q=q$?



Generally speaking, should not $#P=p^n$ and $#Q=q^m$ for $n,m in mathbbN$ be also possible?



Thx




group-theory sylow-theory




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  • 3




    The order of $P$ divides the order of $G$...
    – dan_fulea
    Jul 27 at 16:31






  • 2




    Lagrange's Theorem (group theory)
    – Clip
    Jul 27 at 16:34














up vote
1
down vote

favorite












Let $G$ be a group with $#G=p cdot q$ for $p,q in mathbbP$.



W.l.o.g. $p>q$. According to Sylow theorems we have one Sylow p-subgroup, let's call it $P$. Let $Q$ be a Sylow q-subgroup.



How can I prove that $#P=p$ and $#Q=q$?



Generally speaking, should not $#P=p^n$ and $#Q=q^m$ for $n,m in mathbbN$ be also possible?



Thx




group-theory sylow-theory




share|cite|improve this question

















  • 3




    The order of $P$ divides the order of $G$...
    – dan_fulea
    Jul 27 at 16:31






  • 2




    Lagrange's Theorem (group theory)
    – Clip
    Jul 27 at 16:34












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $G$ be a group with $#G=p cdot q$ for $p,q in mathbbP$.



W.l.o.g. $p>q$. According to Sylow theorems we have one Sylow p-subgroup, let's call it $P$. Let $Q$ be a Sylow q-subgroup.



How can I prove that $#P=p$ and $#Q=q$?



Generally speaking, should not $#P=p^n$ and $#Q=q^m$ for $n,m in mathbbN$ be also possible?



Thx




group-theory sylow-theory




share|cite|improve this question













Let $G$ be a group with $#G=p cdot q$ for $p,q in mathbbP$.



W.l.o.g. $p>q$. According to Sylow theorems we have one Sylow p-subgroup, let's call it $P$. Let $Q$ be a Sylow q-subgroup.



How can I prove that $#P=p$ and $#Q=q$?



Generally speaking, should not $#P=p^n$ and $#Q=q^m$ for $n,m in mathbbN$ be also possible?



Thx





group-theory sylow-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 16:32
























asked Jul 27 at 16:30









Clip

64




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  • 3




    The order of $P$ divides the order of $G$...
    – dan_fulea
    Jul 27 at 16:31






  • 2




    Lagrange's Theorem (group theory)
    – Clip
    Jul 27 at 16:34












  • 3




    The order of $P$ divides the order of $G$...
    – dan_fulea
    Jul 27 at 16:31






  • 2




    Lagrange's Theorem (group theory)
    – Clip
    Jul 27 at 16:34







3




3




The order of $P$ divides the order of $G$...
– dan_fulea
Jul 27 at 16:31




The order of $P$ divides the order of $G$...
– dan_fulea
Jul 27 at 16:31




2




2




Lagrange's Theorem (group theory)
– Clip
Jul 27 at 16:34




Lagrange's Theorem (group theory)
– Clip
Jul 27 at 16:34










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By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.



So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.






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    By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.



    So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.






    share|cite|improve this answer

























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      By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.



      So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.






      share|cite|improve this answer























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        up vote
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        down vote









        By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.



        So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.






        share|cite|improve this answer













        By definition, if $p$ is a prime dividing $|G|$, then a Sylow $p$-subgroup of $G$ is of order $p^n$ where $|G|=p^nk$ such that $(p,k)=1$. In other words, the order of a Sylow $p$-subgroup is the highest power of $p$ in the order of $G$.



        So since $|G|=pq$ where $p,q$ are distinct primes, a Sylow $p$-subgroup of $G$ must be of order $p$.







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Jul 28 at 0:53









        Alan Wang

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