Misunderstanding of Sylow theory Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Elementary Group Theory, Sylow TheoremsSylow theorems and normalizerNumber of sylow subgroups of $A_5$ and $S_5$ - Dummit foote $4.5.31$A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup.Proving that a subgroup $|H|=p^k$ is a Sylow subgroup of $|G|=p^km$, $mnmid p$Sylow Theorems to find all groups with order less than or equal to 10Finding and classifying all groups of order 12Question from Herstein's Topics in Algebra on Sylow subgroupsProof verification - the only group of order 24 without normal sylow subgroup is $S_4$.Is this proof on the Sylow 5-subgroups of G 100% correct?
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Misunderstanding of Sylow theory
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Misunderstanding of Sylow theory
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Elementary Group Theory, Sylow TheoremsSylow theorems and normalizerNumber of sylow subgroups of $A_5$ and $S_5$ - Dummit foote $4.5.31$A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup.Proving that a subgroup $|H|=p^k$ is a Sylow subgroup of $|G|=p^km$, $mnmid p$Sylow Theorems to find all groups with order less than or equal to 10Finding and classifying all groups of order 12Question from Herstein's Topics in Algebra on Sylow subgroupsProof verification - the only group of order 24 without normal sylow subgroup is $S_4$.Is this proof on the Sylow 5-subgroups of G 100% correct?
$begingroup$
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
asked 29 mins ago
PerturbativePerturbative
4,54821554
$endgroup$
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
asked 29 mins ago
PerturbativePerturbative
4,54821554
$endgroup$
$begingroup$
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
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago
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
asked 29 mins ago
PerturbativePerturbative
4,54821554
$endgroup$
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
group-theory finite-groups sylow-theory
asked 29 mins ago
PerturbativePerturbative
4,54821554
asked 29 mins ago
PerturbativePerturbative
4,54821554
asked 29 mins ago
PerturbativePerturbative
4,54821554
asked 29 mins ago
PerturbativePerturbative
4,54821554
asked 29 mins ago
PerturbativePerturbative
4,54821554
4,54821554
$begingroup$
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
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago
add a comment |
$begingroup$
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
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago
$begingroup$
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
23 mins ago
$begingroup$
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
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 23 mins ago
MarkMark
11k1723
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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$begingroup$
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.
answered 23 mins ago
MarkMark
11k1723
$endgroup$
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.
answered 23 mins ago
MarkMark
11k1723
$endgroup$
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.
answered 23 mins ago
MarkMark
11k1723
$endgroup$
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.
answered 23 mins ago
MarkMark
11k1723
answered 23 mins ago
MarkMark
11k1723
answered 23 mins ago
MarkMark
11k1723
answered 23 mins ago
MarkMark
11k1723
11k1723
add a comment |
add a comment |
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$begingroup$
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
23 mins ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
23 mins ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
20 mins ago