Mixing
A simple question on groups and quotient spaces.
Consider a group and its subgroup:
$$,K,<,G,;.$$
An element $gin G$ generates a bijection of $,G,$ onto itself,
begin{eqnarray}
hat{cal{L}}_g;colonquad Glongrightarrow G;,qquad
x;longrightarrow;hat{
cal
{L}}_g,x;=;g;x;,,
end{eqnarray}
and a bijection of the coset space onto itself,
begin{eqnarray}
hat{L}_g;colonquad G/Klongrightarrow G/K;,qquad
x,K;longrightarrow;hat{L}_g,(x,K);=;g;x;K;,.
end{eqnarray}
The set of operators $,{mathtt{L}},=,{,hat{mathit{L}}_g;big{|};gin G,},$ can be interpreted as a map
begin{eqnarray}
{mathtt{L}},:quad G longrightarrow operatorname{Aut_{set}}(G/K),;,
end{eqnarray}
where $,operatorname{Aut_{set}}(G/K),$ denotes the set of all bijections of $,G/K,$ onto itself.
Is this map into or onto?
In other words, can any bijection of $,G/K,$ onto itself be written as $,L_g,$, with $,gin G,$?
group-theory
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
add a comment |
Consider a group and its subgroup:
$$,K,<,G,;.$$
An element $gin G$ generates a bijection of $,G,$ onto itself,
begin{eqnarray}
hat{cal{L}}_g;colonquad Glongrightarrow G;,qquad
x;longrightarrow;hat{
cal
{L}}_g,x;=;g;x;,,
end{eqnarray}
and a bijection of the coset space onto itself,
begin{eqnarray}
hat{L}_g;colonquad G/Klongrightarrow G/K;,qquad
x,K;longrightarrow;hat{L}_g,(x,K);=;g;x;K;,.
end{eqnarray}
The set of operators $,{mathtt{L}},=,{,hat{mathit{L}}_g;big{|};gin G,},$ can be interpreted as a map
begin{eqnarray}
{mathtt{L}},:quad G longrightarrow operatorname{Aut_{set}}(G/K),;,
end{eqnarray}
where $,operatorname{Aut_{set}}(G/K),$ denotes the set of all bijections of $,G/K,$ onto itself.
Is this map into or onto?
In other words, can any bijection of $,G/K,$ onto itself be written as $,L_g,$, with $,gin G,$?
group-theory
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
3
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
2
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
1
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06
add a comment |
Consider a group and its subgroup:
$$,K,<,G,;.$$
An element $gin G$ generates a bijection of $,G,$ onto itself,
begin{eqnarray}
hat{cal{L}}_g;colonquad Glongrightarrow G;,qquad
x;longrightarrow;hat{
cal
{L}}_g,x;=;g;x;,,
end{eqnarray}
and a bijection of the coset space onto itself,
begin{eqnarray}
hat{L}_g;colonquad G/Klongrightarrow G/K;,qquad
x,K;longrightarrow;hat{L}_g,(x,K);=;g;x;K;,.
end{eqnarray}
The set of operators $,{mathtt{L}},=,{,hat{mathit{L}}_g;big{|};gin G,},$ can be interpreted as a map
begin{eqnarray}
{mathtt{L}},:quad G longrightarrow operatorname{Aut_{set}}(G/K),;,
end{eqnarray}
where $,operatorname{Aut_{set}}(G/K),$ denotes the set of all bijections of $,G/K,$ onto itself.
Is this map into or onto?
In other words, can any bijection of $,G/K,$ onto itself be written as $,L_g,$, with $,gin G,$?
group-theory
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
Consider a group and its subgroup:
$$,K,<,G,;.$$
An element $gin G$ generates a bijection of $,G,$ onto itself,
begin{eqnarray}
hat{cal{L}}_g;colonquad Glongrightarrow G;,qquad
x;longrightarrow;hat{
cal
{L}}_g,x;=;g;x;,,
end{eqnarray}
and a bijection of the coset space onto itself,
begin{eqnarray}
hat{L}_g;colonquad G/Klongrightarrow G/K;,qquad
x,K;longrightarrow;hat{L}_g,(x,K);=;g;x;K;,.
end{eqnarray}
The set of operators $,{mathtt{L}},=,{,hat{mathit{L}}_g;big{|};gin G,},$ can be interpreted as a map
begin{eqnarray}
{mathtt{L}},:quad G longrightarrow operatorname{Aut_{set}}(G/K),;,
end{eqnarray}
where $,operatorname{Aut_{set}}(G/K),$ denotes the set of all bijections of $,G/K,$ onto itself.
Is this map into or onto?
In other words, can any bijection of $,G/K,$ onto itself be written as $,L_g,$, with $,gin G,$?
group-theory
group-theory
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
asked Nov 22 '18 at 2:32
Michael_1812Michael_1812
1409
1409
3
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
2
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
1
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06
add a comment |
3
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
2
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
1
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06
3
3
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
2
2
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
1
1
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06
add a comment |
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3
There are two extreme cases here, $K = { e }$ and $K = G$. The first case shows that $L$ is not surjective in general and the second case shows that $L$ is not injective in general either.
– Qiaochu Yuan
Nov 22 '18 at 2:45
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$
– Michael_1812
Nov 22 '18 at 2:51
2
Yes. And there are $|G|$ of these maps, while there are $|G|!$ maps from $G$ to itself.
– Qiaochu Yuan
Nov 22 '18 at 2:53
If $K={e}$ then $G/K$ coincides with $G$. Accordingly, the left translation on the coset space coincides with a corresponding left translation on the group: $$L_g = {cal L}_g,;. $$ You say that the so-obtained map $Glongrightarrow$ Aut$_{set}(G/K)$ is not surjective. Thereby you imply that there exist bijections of $G/K$ (in our case, bijections of $G$) which are not $cal{L}_g$. What kind of bijections are these? Is this, say, $glongrightarrow g^{-1}~$ ?
– Michael_1812
Nov 22 '18 at 2:57
1
The bijections are not any particular kind. Almost all bijections of $G$ are not left translations: for example, in general the bijection which simply exchanges two elements of $G$.
– Hew Wolff
Nov 22 '18 at 3:06