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
2,517522
$endgroup$
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
2,517522
$endgroup$
$begingroup$
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
2,517522
$endgroup$
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
2,517522
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
2,517522
edited Dec 24 '18 at 15:42
Dbchatto67
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
2,517522
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
2,517522
asked Dec 24 '18 at 15:38
Dbchatto67Dbchatto67
2,517522
2,517522
$begingroup$
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 |
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$begingroup$
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