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Misunderstanding of Sylow theory

6

3

$begingroup$

I think I have a misunderstanding a part of Sylow theory for groups or I have made a big mistake in my reasoning below.

We have the following lemma in Sylow theory:

Let $G$ be a finite group and let $P$ be a Sylow-$p$ subgroup of $G$. If $g in G$ such that the order of $g$ is $p^k$ for some $k$, then $g in P$.

Now consider $S_5$, it has order $120 = 2^3 times 3 times 5$. Take a Sylow-$2$ subgroup $H$ of $S_5$. Sylow theory tells us that $|H| = 2^3 = 8$. Now any $4$-cycle in $S_5$ has order $4 = 2^2$. So any $4$-cycle in $S_5$ must be contained in $H$. But the number of $4$-cycles in $S_5$ is $frac5!(5-4)!4 = frac1204 = 30$, so $S_5$ has at least $30$ elements of order $4$ all of which must be contained in $H$ which has order $8$, an obvious contradiction.

What have I done wrong here?

group-theory finite-groups sylow-theory

share|cite|improve this question

asked yesterday

PerturbativePerturbative

4,55821554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The lemma is just not right. One thing that would be true is that there exists $hin G$ such that $g^hin P$. Are you sure you have all the hypotheses of the Lemma down?
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
  $endgroup$
  – Captain Lama
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think your Sylow p-subgroup $P$ is unique?
  $endgroup$
  – Anton Zagrivin
  yesterday
  
  

  
  
  
  

add a comment | 

6

3

$begingroup$

I think I have a misunderstanding a part of Sylow theory for groups or I have made a big mistake in my reasoning below.

We have the following lemma in Sylow theory:

Let $G$ be a finite group and let $P$ be a Sylow-$p$ subgroup of $G$. If $g in G$ such that the order of $g$ is $p^k$ for some $k$, then $g in P$.

Now consider $S_5$, it has order $120 = 2^3 times 3 times 5$. Take a Sylow-$2$ subgroup $H$ of $S_5$. Sylow theory tells us that $|H| = 2^3 = 8$. Now any $4$-cycle in $S_5$ has order $4 = 2^2$. So any $4$-cycle in $S_5$ must be contained in $H$. But the number of $4$-cycles in $S_5$ is $frac5!(5-4)!4 = frac1204 = 30$, so $S_5$ has at least $30$ elements of order $4$ all of which must be contained in $H$ which has order $8$, an obvious contradiction.

What have I done wrong here?

group-theory finite-groups sylow-theory

share|cite|improve this question

asked yesterday

PerturbativePerturbative

4,55821554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The lemma is just not right. One thing that would be true is that there exists $hin G$ such that $g^hin P$. Are you sure you have all the hypotheses of the Lemma down?
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
  $endgroup$
  – Captain Lama
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think your Sylow p-subgroup $P$ is unique?
  $endgroup$
  – Anton Zagrivin
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

I think I have a misunderstanding a part of Sylow theory for groups or I have made a big mistake in my reasoning below.

We have the following lemma in Sylow theory:

Let $G$ be a finite group and let $P$ be a Sylow-$p$ subgroup of $G$. If $g in G$ such that the order of $g$ is $p^k$ for some $k$, then $g in P$.

Now consider $S_5$, it has order $120 = 2^3 times 3 times 5$. Take a Sylow-$2$ subgroup $H$ of $S_5$. Sylow theory tells us that $|H| = 2^3 = 8$. Now any $4$-cycle in $S_5$ has order $4 = 2^2$. So any $4$-cycle in $S_5$ must be contained in $H$. But the number of $4$-cycles in $S_5$ is $frac5!(5-4)!4 = frac1204 = 30$, so $S_5$ has at least $30$ elements of order $4$ all of which must be contained in $H$ which has order $8$, an obvious contradiction.

What have I done wrong here?

group-theory finite-groups sylow-theory

share|cite|improve this question

asked yesterday

PerturbativePerturbative

4,55821554

$endgroup$

share|cite|improve this question

asked yesterday

PerturbativePerturbative

4,55821554

share|cite|improve this question

asked yesterday

PerturbativePerturbative

4,55821554

asked yesterday

PerturbativePerturbative

4,55821554

asked yesterday

PerturbativePerturbative

4,55821554

1 Answer
1

active

oldest

votes

13

$begingroup$

The lemma you stated is false. It should be like this: if $gin G$ has order $p^k$ for some $kinmathbbN$ then there exists a $p$-Sylow subgroup $Pleq G$ such that $gin P$. So it doesn't say that such an element $g$ must be in all $p$-Sylow subgroups of $G$. The version that you wrote is true if there is only one $p$-Sylow subgroup.

share|cite|improve this answer

answered yesterday

MarkMark

11.1k1824

$endgroup$

add a comment | 

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13

$begingroup$

The lemma you stated is false. It should be like this: if $gin G$ has order $p^k$ for some $kinmathbbN$ then there exists a $p$-Sylow subgroup $Pleq G$ such that $gin P$. So it doesn't say that such an element $g$ must be in all $p$-Sylow subgroups of $G$. The version that you wrote is true if there is only one $p$-Sylow subgroup.

share|cite|improve this answer

answered yesterday

MarkMark

11.1k1824

$endgroup$

add a comment | 

13

$begingroup$

The lemma you stated is false. It should be like this: if $gin G$ has order $p^k$ for some $kinmathbbN$ then there exists a $p$-Sylow subgroup $Pleq G$ such that $gin P$. So it doesn't say that such an element $g$ must be in all $p$-Sylow subgroups of $G$. The version that you wrote is true if there is only one $p$-Sylow subgroup.

share|cite|improve this answer

answered yesterday

MarkMark

11.1k1824

$endgroup$

add a comment | 

$begingroup$

The lemma you stated is false. It should be like this: if $gin G$ has order $p^k$ for some $kinmathbbN$ then there exists a $p$-Sylow subgroup $Pleq G$ such that $gin P$. So it doesn't say that such an element $g$ must be in all $p$-Sylow subgroups of $G$. The version that you wrote is true if there is only one $p$-Sylow subgroup.

share|cite|improve this answer

answered yesterday

MarkMark

11.1k1824

$endgroup$

share|cite|improve this answer

answered yesterday

MarkMark

11.1k1824

answered yesterday

MarkMark

11.1k1824

answered yesterday

MarkMark

11.1k1824