Normal subgroups orbits

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












Let $G$ be a topological group acting transitively and effectively on the space $X$ and let $J,K$ be two normal subgroups of $G$ such that $G=Jcdot K$ and $Jcap Knot =e$. Let $Gx_0$ be the orbit of $x_0in X$ with isotropy subgroup $H$, can we decompose this orbit by $J$ and $K$ orbits, for example something like $G/H=J/(Jcap H)times K/(Kcap H)$? What about the intersection $Jcap K$ action?




abstract-algebra group-theory topological-groups




share|cite|improve this question

















  • 1




    if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
    – Christopher
    Jul 26 at 8:47










  • No, maybe the action is not free
    – Ronald
    Jul 26 at 8:48










  • effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
    – Ronald
    Jul 26 at 8:52







  • 1




    I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
    – Derek Holt
    Jul 26 at 8:53






  • 1




    Yes, but not $Gx_0=G$!
    – Ronald
    Jul 26 at 8:56














up vote
3
down vote

favorite












Let $G$ be a topological group acting transitively and effectively on the space $X$ and let $J,K$ be two normal subgroups of $G$ such that $G=Jcdot K$ and $Jcap Knot =e$. Let $Gx_0$ be the orbit of $x_0in X$ with isotropy subgroup $H$, can we decompose this orbit by $J$ and $K$ orbits, for example something like $G/H=J/(Jcap H)times K/(Kcap H)$? What about the intersection $Jcap K$ action?




abstract-algebra group-theory topological-groups




share|cite|improve this question

















  • 1




    if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
    – Christopher
    Jul 26 at 8:47










  • No, maybe the action is not free
    – Ronald
    Jul 26 at 8:48










  • effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
    – Ronald
    Jul 26 at 8:52







  • 1




    I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
    – Derek Holt
    Jul 26 at 8:53






  • 1




    Yes, but not $Gx_0=G$!
    – Ronald
    Jul 26 at 8:56












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $G$ be a topological group acting transitively and effectively on the space $X$ and let $J,K$ be two normal subgroups of $G$ such that $G=Jcdot K$ and $Jcap Knot =e$. Let $Gx_0$ be the orbit of $x_0in X$ with isotropy subgroup $H$, can we decompose this orbit by $J$ and $K$ orbits, for example something like $G/H=J/(Jcap H)times K/(Kcap H)$? What about the intersection $Jcap K$ action?




abstract-algebra group-theory topological-groups




share|cite|improve this question













Let $G$ be a topological group acting transitively and effectively on the space $X$ and let $J,K$ be two normal subgroups of $G$ such that $G=Jcdot K$ and $Jcap Knot =e$. Let $Gx_0$ be the orbit of $x_0in X$ with isotropy subgroup $H$, can we decompose this orbit by $J$ and $K$ orbits, for example something like $G/H=J/(Jcap H)times K/(Kcap H)$? What about the intersection $Jcap K$ action?





abstract-algebra group-theory topological-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 8:49
























asked Jul 26 at 8:30









Ronald

1,5841821




1,5841821







  • 1




    if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
    – Christopher
    Jul 26 at 8:47










  • No, maybe the action is not free
    – Ronald
    Jul 26 at 8:48










  • effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
    – Ronald
    Jul 26 at 8:52







  • 1




    I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
    – Derek Holt
    Jul 26 at 8:53






  • 1




    Yes, but not $Gx_0=G$!
    – Ronald
    Jul 26 at 8:56












  • 1




    if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
    – Christopher
    Jul 26 at 8:47










  • No, maybe the action is not free
    – Ronald
    Jul 26 at 8:48










  • effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
    – Ronald
    Jul 26 at 8:52







  • 1




    I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
    – Derek Holt
    Jul 26 at 8:53






  • 1




    Yes, but not $Gx_0=G$!
    – Ronald
    Jul 26 at 8:56







1




1




if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
– Christopher
Jul 26 at 8:47




if $G$ acts transitively on $X$ then doesn't $Gx_0 = G$?
– Christopher
Jul 26 at 8:47












No, maybe the action is not free
– Ronald
Jul 26 at 8:48




No, maybe the action is not free
– Ronald
Jul 26 at 8:48












effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
– Ronald
Jul 26 at 8:52





effectively means if $g_1x=x$ for all $xin X$ then $g_1=e$ and isotropy means the stabilizer of $x_0$ which is $gin G; gx_0=x_0$
– Ronald
Jul 26 at 8:52





1




1




I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
– Derek Holt
Jul 26 at 8:53




I guess effectively means faithfully, and isotropy subgroup means stabilizer. But if $G$ acts transitively then $Gx_0 = X$.
– Derek Holt
Jul 26 at 8:53




1




1




Yes, but not $Gx_0=G$!
– Ronald
Jul 26 at 8:56




Yes, but not $Gx_0=G$!
– Ronald
Jul 26 at 8:56















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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863212%2fnormal-subgroups-orbits%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863212%2fnormal-subgroups-orbits%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more