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
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 1 hour 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 1 hour 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
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour 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 1 hour 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 1 hour ago
PerturbativePerturbative
4,54821554
asked 1 hour ago
PerturbativePerturbative
4,54821554
asked 1 hour ago
PerturbativePerturbative
4,54821554
asked 1 hour ago
PerturbativePerturbative
4,54821554
asked 1 hour 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
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour 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
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour 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
1 hour 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
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour 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 1 hour ago
MarkMark
11k1724
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
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 1 hour ago
MarkMark
11k1724
$endgroup$
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 1 hour ago
MarkMark
11k1724
$endgroup$
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 1 hour ago
MarkMark
11k1724
$endgroup$
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 1 hour ago
MarkMark
11k1724
answered 1 hour ago
MarkMark
11k1724
answered 1 hour ago
MarkMark
11k1724
answered 1 hour ago
MarkMark
11k1724
11k1724
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
1 hour ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
1 hour ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
1 hour ago