Finite subgroups of motoins always fix some point on the plane.
$begingroup$
Let $G$ be a finite subgroup of the group of motoins $M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $G$.
The proof of this is sketched by our instructor as follows $:$
Let $s$ be an element on the plane. Let $S$ be the set of all images of $s$ under the action of the elements of $G$ on $s$. Then every element of $G$ permutes the elements of $S$. To see this let us take $f in G,s' in S$. If we can show that $f(s') in S$ then we are through. Since $s' in S$ $exists$ $g in G$ such that $g(s) = s'$. Since $f,g in G$ so $fg in G$. Therefore $fg (s) in G$ i.e. $f(s') in S,$ as claimed. Since $G$ is finite, $S$ is also finite. Let $S = {s_1,s_2, cdots , s_n }.$
After that our instructor asserted that the element $s^{*} = frac {1} {n} (s_1 + s_2 + cdots + s_n)$ is fixed by every element of $G$. This is the stage where I am struggling. Why $s^{*}$ is fixed by every element of $G$? If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $s^{*}$ and permute the $s_i$'s. Since vector addition is commutative we get $s^{*}$ back by acting them on $s^{*}$. Also clearly $G$ doesn't contain any proper translation since they are of infinite order. So what are the elements of $G$ other than rotations about origin and translations about a line through origin? I know that the elements of the group of motoins $M$ are either of the form $t_{a} rho_{theta}$ (which are precisely rotations about some point fixed by it ) or of the form $t_{a} rho_{theta}r$ (which are precisely glide reflections). So according to me the elements of $G$ is either of the form $t_{a} rho_{theta}$ or of the form $rho_{theta} r$. Do they all fix $s^{*}$? Please help me in this regard.
Thank you very much.
abstract-algebra rigid-transformation
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite subgroup of the group of motoins $M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $G$.
The proof of this is sketched by our instructor as follows $:$
Let $s$ be an element on the plane. Let $S$ be the set of all images of $s$ under the action of the elements of $G$ on $s$. Then every element of $G$ permutes the elements of $S$. To see this let us take $f in G,s' in S$. If we can show that $f(s') in S$ then we are through. Since $s' in S$ $exists$ $g in G$ such that $g(s) = s'$. Since $f,g in G$ so $fg in G$. Therefore $fg (s) in G$ i.e. $f(s') in S,$ as claimed. Since $G$ is finite, $S$ is also finite. Let $S = {s_1,s_2, cdots , s_n }.$
After that our instructor asserted that the element $s^{*} = frac {1} {n} (s_1 + s_2 + cdots + s_n)$ is fixed by every element of $G$. This is the stage where I am struggling. Why $s^{*}$ is fixed by every element of $G$? If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $s^{*}$ and permute the $s_i$'s. Since vector addition is commutative we get $s^{*}$ back by acting them on $s^{*}$. Also clearly $G$ doesn't contain any proper translation since they are of infinite order. So what are the elements of $G$ other than rotations about origin and translations about a line through origin? I know that the elements of the group of motoins $M$ are either of the form $t_{a} rho_{theta}$ (which are precisely rotations about some point fixed by it ) or of the form $t_{a} rho_{theta}r$ (which are precisely glide reflections). So according to me the elements of $G$ is either of the form $t_{a} rho_{theta}$ or of the form $rho_{theta} r$. Do they all fix $s^{*}$? Please help me in this regard.
Thank you very much.
abstract-algebra rigid-transformation
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
$endgroup$
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11
add a comment |
$begingroup$
Let $G$ be a finite subgroup of the group of motoins $M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $G$.
The proof of this is sketched by our instructor as follows $:$
Let $s$ be an element on the plane. Let $S$ be the set of all images of $s$ under the action of the elements of $G$ on $s$. Then every element of $G$ permutes the elements of $S$. To see this let us take $f in G,s' in S$. If we can show that $f(s') in S$ then we are through. Since $s' in S$ $exists$ $g in G$ such that $g(s) = s'$. Since $f,g in G$ so $fg in G$. Therefore $fg (s) in G$ i.e. $f(s') in S,$ as claimed. Since $G$ is finite, $S$ is also finite. Let $S = {s_1,s_2, cdots , s_n }.$
After that our instructor asserted that the element $s^{*} = frac {1} {n} (s_1 + s_2 + cdots + s_n)$ is fixed by every element of $G$. This is the stage where I am struggling. Why $s^{*}$ is fixed by every element of $G$? If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $s^{*}$ and permute the $s_i$'s. Since vector addition is commutative we get $s^{*}$ back by acting them on $s^{*}$. Also clearly $G$ doesn't contain any proper translation since they are of infinite order. So what are the elements of $G$ other than rotations about origin and translations about a line through origin? I know that the elements of the group of motoins $M$ are either of the form $t_{a} rho_{theta}$ (which are precisely rotations about some point fixed by it ) or of the form $t_{a} rho_{theta}r$ (which are precisely glide reflections). So according to me the elements of $G$ is either of the form $t_{a} rho_{theta}$ or of the form $rho_{theta} r$. Do they all fix $s^{*}$? Please help me in this regard.
Thank you very much.
abstract-algebra rigid-transformation
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
$endgroup$
Let $G$ be a finite subgroup of the group of motoins $M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $G$.
The proof of this is sketched by our instructor as follows $:$
Let $s$ be an element on the plane. Let $S$ be the set of all images of $s$ under the action of the elements of $G$ on $s$. Then every element of $G$ permutes the elements of $S$. To see this let us take $f in G,s' in S$. If we can show that $f(s') in S$ then we are through. Since $s' in S$ $exists$ $g in G$ such that $g(s) = s'$. Since $f,g in G$ so $fg in G$. Therefore $fg (s) in G$ i.e. $f(s') in S,$ as claimed. Since $G$ is finite, $S$ is also finite. Let $S = {s_1,s_2, cdots , s_n }.$
After that our instructor asserted that the element $s^{*} = frac {1} {n} (s_1 + s_2 + cdots + s_n)$ is fixed by every element of $G$. This is the stage where I am struggling. Why $s^{*}$ is fixed by every element of $G$? If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $s^{*}$ and permute the $s_i$'s. Since vector addition is commutative we get $s^{*}$ back by acting them on $s^{*}$. Also clearly $G$ doesn't contain any proper translation since they are of infinite order. So what are the elements of $G$ other than rotations about origin and translations about a line through origin? I know that the elements of the group of motoins $M$ are either of the form $t_{a} rho_{theta}$ (which are precisely rotations about some point fixed by it ) or of the form $t_{a} rho_{theta}r$ (which are precisely glide reflections). So according to me the elements of $G$ is either of the form $t_{a} rho_{theta}$ or of the form $rho_{theta} r$. Do they all fix $s^{*}$? Please help me in this regard.
Thank you very much.
abstract-algebra rigid-transformation
abstract-algebra rigid-transformation
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
edited Dec 24 '18 at 15:42
Dbchatto67
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
1,629219
1,629219
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11
add a comment |
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11
add a comment |
0
active
oldest
votes
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%2f3051369%2ffinite-subgroups-of-motoins-always-fix-some-point-on-the-plane%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%2f3051369%2ffinite-subgroups-of-motoins-always-fix-some-point-on-the-plane%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
$begingroup$
I have proved it on my own at least. Thank you very much.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:01
$begingroup$
I have proved that any rigid transformations take center of gravity to the center of gravity.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:03
$begingroup$
To be explicit let $m in M$ and let $S={s_1,s_2, cdots , s_n }$ be any finite collection of points on the plane. Let $s = frac {1} {n} (s_1+ s_2 + cdots + s_n)$ then $m(s) = s'$ where $s' = frac {1} {n} (m(s_1) + m(s_2) + cdots + m(s_n))$.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:09
$begingroup$
In this case $m in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:11