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?
What is an "asse" in Elizabethan English?
How does a spellshard spellbook work?
Misunderstanding of Sylow theory
Semigroups with no morphisms between them
What does 丫 mean? 丫是什么意思？
Google .dev domain strangely redirects to https
What is the home of the drow in Flanaess?
How could we fake a moon landing now?
Antipodal Land Area Calculation
What does this say in Elvish?
Why we try to capture variability?
What is the meaning of 'breadth' in breadth first search?
What would you call this weird metallic apparatus that allows you to lift people?
How to run automated tests after each commit?
If Windows 7 doesn't support WSL, then what is "Subsystem for UNIX-based Applications"?
Is CEO the "profession" with the most psychopaths?
How do living politicians protect their readily obtainable signatures from misuse?
One-one communication
Co-worker has annoying ringtone
Drawing spherical mirrors
How often does castling occur in grandmaster games?
Intuitive explanation of the rank-nullity theorem
How does the math work when buying airline miles?
What makes a man succeed?
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 3 hours ago
PerturbativePerturbative
4,55321554
$endgroup$
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 3 hours ago
PerturbativePerturbative
4,55321554
$endgroup$
$begingroup$
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
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours 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 3 hours ago
PerturbativePerturbative
4,55321554
$endgroup$
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 3 hours ago
PerturbativePerturbative
4,55321554
asked 3 hours ago
PerturbativePerturbative
4,55321554
asked 3 hours ago
PerturbativePerturbative
4,55321554
asked 3 hours ago
PerturbativePerturbative
4,55321554
asked 3 hours ago
PerturbativePerturbative
4,55321554
4,55321554
$begingroup$
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
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours 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
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours 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
3 hours 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
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours 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 3 hours ago
MarkMark
11.1k1724
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193946%2fmisunderstanding-of-sylow-theory%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 3 hours ago
MarkMark
11.1k1724
$endgroup$
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 3 hours ago
MarkMark
11.1k1724
$endgroup$
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 3 hours ago
MarkMark
11.1k1724
$endgroup$
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 3 hours ago
MarkMark
11.1k1724
answered 3 hours ago
MarkMark
11.1k1724
answered 3 hours ago
MarkMark
11.1k1724
answered 3 hours ago
MarkMark
11.1k1724
11.1k1724
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193946%2fmisunderstanding-of-sylow-theory%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
3 hours ago
$begingroup$
The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements.
$endgroup$
– Captain Lama
3 hours ago
$begingroup$
Why do you think your Sylow p-subgroup $P$ is unique?
$endgroup$
– Anton Zagrivin
3 hours ago