trying to understand this proof of sylows theorem that say the number of p-sylow subgroups is 1+kp
$begingroup$
I'm Very confused as to what my lecturer means in the final few lines of a proof of one of sylows theorems means. The theorem in question is the one that says the number of sylow P-subgroups is 1+kp.
Here's the part of the proof I'm confused about :
Let S be the set of all subsets of order $p^alpha$ let $Ain S$ and let G act on it by right multiplication
.
.
.
It follows then that the number of P-sylow subgroups is the number of orbits of S of size m . let $n_p$ be the number of p sylow subgroups then
$$
begin{aligned}
& |S|=sum_{text{orbits of size m}}|O|+sum_{text{orbits size >m}}|O|\
Rightarrow & |S|=n_pm+mpr\ Rightarrow & binom{mp^alpha}{p^alpha}/m=n_p+pr \ Rightarrow & binom{mp^alpha}{p^alpha}/mequiv n_ppmod p
end{aligned}$$
But we know that the cyclic group of order $mp^{alpha}$ has exactly one subgroup of order $p^{alpha}$ Thus $ binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
I understodd the rest of the proof just fine , but this has really stumped me. for starters I'm not familiar with the notation he seems to be using to describe the size of S( the brackets term). secondly I don't understand how he made the jump $binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
Could anyone plese help me to understand this ?
Note: the dots represent part of the proof I left out if needs be I can add it for context but I thought it seemed superfluous given that this little part is the source of all my confusion.
group-theory proof-explanation sylow-theory
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
$endgroup$
|
show 7 more comments
$begingroup$
I'm Very confused as to what my lecturer means in the final few lines of a proof of one of sylows theorems means. The theorem in question is the one that says the number of sylow P-subgroups is 1+kp.
Here's the part of the proof I'm confused about :
Let S be the set of all subsets of order $p^alpha$ let $Ain S$ and let G act on it by right multiplication
.
.
.
It follows then that the number of P-sylow subgroups is the number of orbits of S of size m . let $n_p$ be the number of p sylow subgroups then
$$
begin{aligned}
& |S|=sum_{text{orbits of size m}}|O|+sum_{text{orbits size >m}}|O|\
Rightarrow & |S|=n_pm+mpr\ Rightarrow & binom{mp^alpha}{p^alpha}/m=n_p+pr \ Rightarrow & binom{mp^alpha}{p^alpha}/mequiv n_ppmod p
end{aligned}$$
But we know that the cyclic group of order $mp^{alpha}$ has exactly one subgroup of order $p^{alpha}$ Thus $ binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
I understodd the rest of the proof just fine , but this has really stumped me. for starters I'm not familiar with the notation he seems to be using to describe the size of S( the brackets term). secondly I don't understand how he made the jump $binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
Could anyone plese help me to understand this ?
Note: the dots represent part of the proof I left out if needs be I can add it for context but I thought it seemed superfluous given that this little part is the source of all my confusion.
group-theory proof-explanation sylow-theory
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
$endgroup$
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
1
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
1
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
1
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43
|
show 7 more comments
$begingroup$
I'm Very confused as to what my lecturer means in the final few lines of a proof of one of sylows theorems means. The theorem in question is the one that says the number of sylow P-subgroups is 1+kp.
Here's the part of the proof I'm confused about :
Let S be the set of all subsets of order $p^alpha$ let $Ain S$ and let G act on it by right multiplication
.
.
.
It follows then that the number of P-sylow subgroups is the number of orbits of S of size m . let $n_p$ be the number of p sylow subgroups then
$$
begin{aligned}
& |S|=sum_{text{orbits of size m}}|O|+sum_{text{orbits size >m}}|O|\
Rightarrow & |S|=n_pm+mpr\ Rightarrow & binom{mp^alpha}{p^alpha}/m=n_p+pr \ Rightarrow & binom{mp^alpha}{p^alpha}/mequiv n_ppmod p
end{aligned}$$
But we know that the cyclic group of order $mp^{alpha}$ has exactly one subgroup of order $p^{alpha}$ Thus $ binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
I understodd the rest of the proof just fine , but this has really stumped me. for starters I'm not familiar with the notation he seems to be using to describe the size of S( the brackets term). secondly I don't understand how he made the jump $binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
Could anyone plese help me to understand this ?
Note: the dots represent part of the proof I left out if needs be I can add it for context but I thought it seemed superfluous given that this little part is the source of all my confusion.
group-theory proof-explanation sylow-theory
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
$endgroup$
I'm Very confused as to what my lecturer means in the final few lines of a proof of one of sylows theorems means. The theorem in question is the one that says the number of sylow P-subgroups is 1+kp.
Here's the part of the proof I'm confused about :
Let S be the set of all subsets of order $p^alpha$ let $Ain S$ and let G act on it by right multiplication
.
.
.
It follows then that the number of P-sylow subgroups is the number of orbits of S of size m . let $n_p$ be the number of p sylow subgroups then
$$
begin{aligned}
& |S|=sum_{text{orbits of size m}}|O|+sum_{text{orbits size >m}}|O|\
Rightarrow & |S|=n_pm+mpr\ Rightarrow & binom{mp^alpha}{p^alpha}/m=n_p+pr \ Rightarrow & binom{mp^alpha}{p^alpha}/mequiv n_ppmod p
end{aligned}$$
But we know that the cyclic group of order $mp^{alpha}$ has exactly one subgroup of order $p^{alpha}$ Thus $ binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
I understodd the rest of the proof just fine , but this has really stumped me. for starters I'm not familiar with the notation he seems to be using to describe the size of S( the brackets term). secondly I don't understand how he made the jump $binom{mp^alpha}{p^alpha}/mequiv1pmod p$ and so $n_pequiv1pmod p$
Could anyone plese help me to understand this ?
Note: the dots represent part of the proof I left out if needs be I can add it for context but I thought it seemed superfluous given that this little part is the source of all my confusion.
group-theory proof-explanation sylow-theory
group-theory proof-explanation sylow-theory
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
edited Dec 23 '18 at 12:15
quid♦
37.2k95193
37.2k95193
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
asked Dec 23 '18 at 2:03
can'tcauchycan'tcauchy
1,016417
1,016417
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
1
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
1
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
1
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43
|
show 7 more comments
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
1
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
1
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
1
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
1
1
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
1
1
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
1
1
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43
|
show 7 more comments
0
active
oldest
votes
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3050008%2ftrying-to-understand-this-proof-of-sylows-theorem-that-say-the-number-of-p-sylow%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3050008%2ftrying-to-understand-this-proof-of-sylows-theorem-that-say-the-number-of-p-sylow%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
$begingroup$
Not sure I can help (yet), but I'm curious how $m$ is defined.
$endgroup$
– Chris Custer
Dec 23 '18 at 3:51
1
$begingroup$
When you say "bracket term" do you mean $binom{mp^a}{p^s}$? That's a binomial coefficient.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
1
$begingroup$
@ChrisCuster No doubt the order of $G$ is $mp^a$ where $pnmid m$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:01
$begingroup$
Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown
$endgroup$
– Chris Custer
Dec 23 '18 at 5:07
1
$begingroup$
The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up.
$endgroup$
– Chris Custer
Dec 23 '18 at 16:43