Mixing
How to show a certain group element must belong to the stabilizer of a set element
$begingroup$
I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:
Let a group $G$ acting on $Omega$ and take $alpha,betainOmega$, $xin G$ and $alpha^x=beta$ (meaning $x$ acts on $alpha$ to get $beta$).
Show that if $yin G$ and $alpha^y=beta$ $implies exists gin operatorname{Stab}(alpha)$ s.t. $y=gx$, i.e. $yin operatorname{Stab}(alpha)x$.
In words, this is tantamount to claim all group elements moving $alpha$ to $beta$ belong to the group set given by the stabilizer elements of $alpha$ followed by a guess element moving $alpha$ to $beta$.
To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.
That said, I've tried to show that such $g$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete.
It goes like this:
Let's take the element $gin G$ s.t. $y=gx$, i.e. $g=yx^{-1}$. We can also claim $alpha^y=alpha^x$ so, replacing the y: $$alpha^{gx}=alpha^x implies (alpha^g)^x = alpha^x$$
But this can be true only iff:$$alpha^g=alpha implies gin operatorname{Stab}(alpha) implies yin operatorname{Stab}(alpha)x$$
I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $in operatorname{Stab}(alpha)x$ are the only ones bringing $alpha$ to $beta$?
I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $g notin operatorname{Stab}(alpha)$ . . . " and deriving from it a contradiction.
Thanks for your precious support in advance.
group-theory group-actions
edited Jan 12 at 17:04
Shaun
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
$endgroup$
add a comment |
$begingroup$
I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:
Let a group $G$ acting on $Omega$ and take $alpha,betainOmega$, $xin G$ and $alpha^x=beta$ (meaning $x$ acts on $alpha$ to get $beta$).
Show that if $yin G$ and $alpha^y=beta$ $implies exists gin operatorname{Stab}(alpha)$ s.t. $y=gx$, i.e. $yin operatorname{Stab}(alpha)x$.
In words, this is tantamount to claim all group elements moving $alpha$ to $beta$ belong to the group set given by the stabilizer elements of $alpha$ followed by a guess element moving $alpha$ to $beta$.
To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.
That said, I've tried to show that such $g$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete.
It goes like this:
Let's take the element $gin G$ s.t. $y=gx$, i.e. $g=yx^{-1}$. We can also claim $alpha^y=alpha^x$ so, replacing the y: $$alpha^{gx}=alpha^x implies (alpha^g)^x = alpha^x$$
But this can be true only iff:$$alpha^g=alpha implies gin operatorname{Stab}(alpha) implies yin operatorname{Stab}(alpha)x$$
I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $in operatorname{Stab}(alpha)x$ are the only ones bringing $alpha$ to $beta$?
I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $g notin operatorname{Stab}(alpha)$ . . . " and deriving from it a contradiction.
Thanks for your precious support in advance.
group-theory group-actions
edited Jan 12 at 17:04
Shaun
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
$endgroup$
2
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
1
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59
add a comment |
$begingroup$
I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:
Let a group $G$ acting on $Omega$ and take $alpha,betainOmega$, $xin G$ and $alpha^x=beta$ (meaning $x$ acts on $alpha$ to get $beta$).
Show that if $yin G$ and $alpha^y=beta$ $implies exists gin operatorname{Stab}(alpha)$ s.t. $y=gx$, i.e. $yin operatorname{Stab}(alpha)x$.
In words, this is tantamount to claim all group elements moving $alpha$ to $beta$ belong to the group set given by the stabilizer elements of $alpha$ followed by a guess element moving $alpha$ to $beta$.
To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.
That said, I've tried to show that such $g$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete.
It goes like this:
Let's take the element $gin G$ s.t. $y=gx$, i.e. $g=yx^{-1}$. We can also claim $alpha^y=alpha^x$ so, replacing the y: $$alpha^{gx}=alpha^x implies (alpha^g)^x = alpha^x$$
But this can be true only iff:$$alpha^g=alpha implies gin operatorname{Stab}(alpha) implies yin operatorname{Stab}(alpha)x$$
I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $in operatorname{Stab}(alpha)x$ are the only ones bringing $alpha$ to $beta$?
I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $g notin operatorname{Stab}(alpha)$ . . . " and deriving from it a contradiction.
Thanks for your precious support in advance.
group-theory group-actions
edited Jan 12 at 17:04
Shaun
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
$endgroup$
I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:
Let a group $G$ acting on $Omega$ and take $alpha,betainOmega$, $xin G$ and $alpha^x=beta$ (meaning $x$ acts on $alpha$ to get $beta$).
Show that if $yin G$ and $alpha^y=beta$ $implies exists gin operatorname{Stab}(alpha)$ s.t. $y=gx$, i.e. $yin operatorname{Stab}(alpha)x$.
In words, this is tantamount to claim all group elements moving $alpha$ to $beta$ belong to the group set given by the stabilizer elements of $alpha$ followed by a guess element moving $alpha$ to $beta$.
To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.
That said, I've tried to show that such $g$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete.
It goes like this:
Let's take the element $gin G$ s.t. $y=gx$, i.e. $g=yx^{-1}$. We can also claim $alpha^y=alpha^x$ so, replacing the y: $$alpha^{gx}=alpha^x implies (alpha^g)^x = alpha^x$$
But this can be true only iff:$$alpha^g=alpha implies gin operatorname{Stab}(alpha) implies yin operatorname{Stab}(alpha)x$$
I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $in operatorname{Stab}(alpha)x$ are the only ones bringing $alpha$ to $beta$?
I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $g notin operatorname{Stab}(alpha)$ . . . " and deriving from it a contradiction.
Thanks for your precious support in advance.
group-theory group-actions
group-theory group-actions
edited Jan 12 at 17:04
Shaun
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
edited Jan 12 at 17:04
Shaun
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
edited Jan 12 at 17:04
Shaun
9,082113683
edited Jan 12 at 17:04
Shaun
9,082113683
edited Jan 12 at 17:04
Shaun
9,082113683
9,082113683
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
asked Jan 12 at 15:23
riccardoventrellariccardoventrella
658
658
2
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
1
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59
add a comment |
2
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
1
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59
2
2
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
1
1
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The proof is enough. Some further clarification:
Claim: $alpha^x = alpha^y implies y in Stab(alpha)x$
Proof: As you noted, there exists $g in G$ such that $y = gx$. Thus,
$$ (alpha^g)^x = alpha^x$$
It suffices to show that $alpha ^ g = alpha$, as this would imply that $y in Stab(alpha)x = { gx | alpha^g = alpha}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((alpha^g)^x)^{x^{-1}} = (alpha^x)^{x^{-1}}$$
so you would get
$$ alpha^g = alpha$$
as desired.
answered Jan 12 at 16:56
MetricMetric
1,22149
$endgroup$
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
add a comment |
Your Answer
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%2f3071005%2fhow-to-show-a-certain-group-element-must-belong-to-the-stabilizer-of-a-set-eleme%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 proof is enough. Some further clarification:
Claim: $alpha^x = alpha^y implies y in Stab(alpha)x$
Proof: As you noted, there exists $g in G$ such that $y = gx$. Thus,
$$ (alpha^g)^x = alpha^x$$
It suffices to show that $alpha ^ g = alpha$, as this would imply that $y in Stab(alpha)x = { gx | alpha^g = alpha}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((alpha^g)^x)^{x^{-1}} = (alpha^x)^{x^{-1}}$$
so you would get
$$ alpha^g = alpha$$
as desired.
answered Jan 12 at 16:56
MetricMetric
1,22149
$endgroup$
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
add a comment |
$begingroup$
The proof is enough. Some further clarification:
Claim: $alpha^x = alpha^y implies y in Stab(alpha)x$
Proof: As you noted, there exists $g in G$ such that $y = gx$. Thus,
$$ (alpha^g)^x = alpha^x$$
It suffices to show that $alpha ^ g = alpha$, as this would imply that $y in Stab(alpha)x = { gx | alpha^g = alpha}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((alpha^g)^x)^{x^{-1}} = (alpha^x)^{x^{-1}}$$
so you would get
$$ alpha^g = alpha$$
as desired.
answered Jan 12 at 16:56
MetricMetric
1,22149
$endgroup$
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
add a comment |
$begingroup$
The proof is enough. Some further clarification:
Claim: $alpha^x = alpha^y implies y in Stab(alpha)x$
Proof: As you noted, there exists $g in G$ such that $y = gx$. Thus,
$$ (alpha^g)^x = alpha^x$$
It suffices to show that $alpha ^ g = alpha$, as this would imply that $y in Stab(alpha)x = { gx | alpha^g = alpha}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((alpha^g)^x)^{x^{-1}} = (alpha^x)^{x^{-1}}$$
so you would get
$$ alpha^g = alpha$$
as desired.
answered Jan 12 at 16:56
MetricMetric
1,22149
$endgroup$
The proof is enough. Some further clarification:
Claim: $alpha^x = alpha^y implies y in Stab(alpha)x$
Proof: As you noted, there exists $g in G$ such that $y = gx$. Thus,
$$ (alpha^g)^x = alpha^x$$
It suffices to show that $alpha ^ g = alpha$, as this would imply that $y in Stab(alpha)x = { gx | alpha^g = alpha}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((alpha^g)^x)^{x^{-1}} = (alpha^x)^{x^{-1}}$$
so you would get
$$ alpha^g = alpha$$
as desired.
answered Jan 12 at 16:56
MetricMetric
1,22149
answered Jan 12 at 16:56
MetricMetric
1,22149
answered Jan 12 at 16:56
MetricMetric
1,22149
answered Jan 12 at 16:56
MetricMetric
1,22149
1,22149
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
add a comment |
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $alpha^g = alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out.
$endgroup$
– riccardoventrella
Jan 12 at 17:04
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
$begingroup$
You're welcome!
$endgroup$
– Metric
Jan 12 at 17:05
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%2f3071005%2fhow-to-show-a-certain-group-element-must-belong-to-the-stabilizer-of-a-set-eleme%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
2
$begingroup$
I'm just assuming $gin G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $gin G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $in Stab(alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $in Stab(alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course.
$endgroup$
– riccardoventrella
Jan 12 at 16:37
$begingroup$
My bad. Couldn't realise it first time.
$endgroup$
– Thomas Shelby
Jan 12 at 16:47
1
$begingroup$
no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $in Stab(alpha)$ indeed (which $e$ belongs to of course, as subgroup)
$endgroup$
– riccardoventrella
Jan 12 at 16:59