Mixing
Free action of a subgroup's normaliser on a set of orbits
$begingroup$
I'm struggling with this question:
Let $G$ be a group that acts freely on a set $X$, and $K < G$. Prove that the normaliser of $K$, $N(K)$, acts freely on $X/K$.
The question was posed to me like this, but it doesn't even seem to be fully specified. How can we prove that $N(K)$ acts freely on $X/K$ when we don't even know how it acts on $X/K$? Do we assume that it acts in the natural way, i.e. that $n *(K cdot x) = (nK)cdot x$? I've made this assumption but keep getting stuck when I try to prove it.
group-theory finite-groups group-actions
asked Jan 27 at 13:47
D GD G
1628
$endgroup$
add a comment |
$begingroup$
I'm struggling with this question:
Let $G$ be a group that acts freely on a set $X$, and $K < G$. Prove that the normaliser of $K$, $N(K)$, acts freely on $X/K$.
The question was posed to me like this, but it doesn't even seem to be fully specified. How can we prove that $N(K)$ acts freely on $X/K$ when we don't even know how it acts on $X/K$? Do we assume that it acts in the natural way, i.e. that $n *(K cdot x) = (nK)cdot x$? I've made this assumption but keep getting stuck when I try to prove it.
group-theory finite-groups group-actions
asked Jan 27 at 13:47
D GD G
1628
$endgroup$
1
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51
add a comment |
$begingroup$
I'm struggling with this question:
Let $G$ be a group that acts freely on a set $X$, and $K < G$. Prove that the normaliser of $K$, $N(K)$, acts freely on $X/K$.
The question was posed to me like this, but it doesn't even seem to be fully specified. How can we prove that $N(K)$ acts freely on $X/K$ when we don't even know how it acts on $X/K$? Do we assume that it acts in the natural way, i.e. that $n *(K cdot x) = (nK)cdot x$? I've made this assumption but keep getting stuck when I try to prove it.
group-theory finite-groups group-actions
asked Jan 27 at 13:47
D GD G
1628
$endgroup$
I'm struggling with this question:
Let $G$ be a group that acts freely on a set $X$, and $K < G$. Prove that the normaliser of $K$, $N(K)$, acts freely on $X/K$.
The question was posed to me like this, but it doesn't even seem to be fully specified. How can we prove that $N(K)$ acts freely on $X/K$ when we don't even know how it acts on $X/K$? Do we assume that it acts in the natural way, i.e. that $n *(K cdot x) = (nK)cdot x$? I've made this assumption but keep getting stuck when I try to prove it.
group-theory finite-groups group-actions
group-theory finite-groups group-actions
asked Jan 27 at 13:47
D GD G
1628
asked Jan 27 at 13:47
D GD G
1628
edited Jan 27 at 14:51
D G
asked Jan 27 at 13:47
D GD G
1628
asked Jan 27 at 13:47
D GD G
1628
asked Jan 27 at 13:47
D GD G
1628
1628
1
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51
add a comment |
1
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51
1
1
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:Xrightarrow X/K$ the quotient map. Let $gin N(K)$ and $xin X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, kin K$, $gy=gkx=gkg^{-1}g(x)$ since $gin N(K)$, $gkg^{-1}in K$ and $p(g.x)=p(g.y)$.
Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $bar gin N(K)/K$ Suppose that $bar g.p(x)=p(x)$. Let $gin N(K)$ whose is image by the quotient map $N(K)rightarrow N(K)/K$ is $bar g$, we deduce that $gx=kx, kin K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
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%2f3089577%2ffree-action-of-a-subgroups-normaliser-on-a-set-of-orbits%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
$begingroup$
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:Xrightarrow X/K$ the quotient map. Let $gin N(K)$ and $xin X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, kin K$, $gy=gkx=gkg^{-1}g(x)$ since $gin N(K)$, $gkg^{-1}in K$ and $p(g.x)=p(g.y)$.
Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $bar gin N(K)/K$ Suppose that $bar g.p(x)=p(x)$. Let $gin N(K)$ whose is image by the quotient map $N(K)rightarrow N(K)/K$ is $bar g$, we deduce that $gx=kx, kin K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
add a comment |
$begingroup$
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:Xrightarrow X/K$ the quotient map. Let $gin N(K)$ and $xin X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, kin K$, $gy=gkx=gkg^{-1}g(x)$ since $gin N(K)$, $gkg^{-1}in K$ and $p(g.x)=p(g.y)$.
Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $bar gin N(K)/K$ Suppose that $bar g.p(x)=p(x)$. Let $gin N(K)$ whose is image by the quotient map $N(K)rightarrow N(K)/K$ is $bar g$, we deduce that $gx=kx, kin K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
add a comment |
$begingroup$
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:Xrightarrow X/K$ the quotient map. Let $gin N(K)$ and $xin X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, kin K$, $gy=gkx=gkg^{-1}g(x)$ since $gin N(K)$, $gkg^{-1}in K$ and $p(g.x)=p(g.y)$.
Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $bar gin N(K)/K$ Suppose that $bar g.p(x)=p(x)$. Let $gin N(K)$ whose is image by the quotient map $N(K)rightarrow N(K)/K$ is $bar g$, we deduce that $gx=kx, kin K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:Xrightarrow X/K$ the quotient map. Let $gin N(K)$ and $xin X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, kin K$, $gy=gkx=gkg^{-1}g(x)$ since $gin N(K)$, $gkg^{-1}in K$ and $p(g.x)=p(g.y)$.
Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $bar gin N(K)/K$ Suppose that $bar g.p(x)=p(x)$. Let $gin N(K)$ whose is image by the quotient map $N(K)rightarrow N(K)/K$ is $bar g$, we deduce that $gx=kx, kin K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
edited Jan 27 at 14:01
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
answered Jan 27 at 13:56
Tsemo AristideTsemo Aristide
59.8k11446
59.8k11446
add a comment |
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%2f3089577%2ffree-action-of-a-subgroups-normaliser-on-a-set-of-orbits%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
1
$begingroup$
Can I point out that this is badly worded ?I don't think that $G$ is intended to be a free group in particular, but $G$ is a group, so $G$ cannot be a group action. You mean something like "Suppose that the group $G$ acts freely on the set $X$". There is a natrual indiuced action of $N_G(K)$ on the orbits of $K$ by $n.(Kx) := K(n.x)$, for $n in N(K)$ and $x in X$.
$endgroup$
– Derek Holt
Jan 27 at 13:55
$begingroup$
fixed the wording
$endgroup$
– D G
Jan 27 at 14:51