Does every subgroup of finite index contain a power of each element of the group?
$begingroup$
Let $G$ be a group, not necessarily finite. If $H$ is a normal subgroup of $G$ of a finite index, say $(G:H)=n$, then for every $gin G$ we have $g^nin H$. Does this statement remain valid if do not assume $H$ to be normal?
In particular let $SL_2(mathbb Z)$ be the modular group, and let $Gammasubset SL_2(mathbb Z)$ be a subgroup of a finite index. Does there exists a positive integer $ell$ such that $begin{pmatrix}1&1\0 & 1end{pmatrix}^ell$ lies in $Gamma$?
group-theory modular-group
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group, not necessarily finite. If $H$ is a normal subgroup of $G$ of a finite index, say $(G:H)=n$, then for every $gin G$ we have $g^nin H$. Does this statement remain valid if do not assume $H$ to be normal?
In particular let $SL_2(mathbb Z)$ be the modular group, and let $Gammasubset SL_2(mathbb Z)$ be a subgroup of a finite index. Does there exists a positive integer $ell$ such that $begin{pmatrix}1&1\0 & 1end{pmatrix}^ell$ lies in $Gamma$?
group-theory modular-group
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
$endgroup$
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26
add a comment |
$begingroup$
Let $G$ be a group, not necessarily finite. If $H$ is a normal subgroup of $G$ of a finite index, say $(G:H)=n$, then for every $gin G$ we have $g^nin H$. Does this statement remain valid if do not assume $H$ to be normal?
In particular let $SL_2(mathbb Z)$ be the modular group, and let $Gammasubset SL_2(mathbb Z)$ be a subgroup of a finite index. Does there exists a positive integer $ell$ such that $begin{pmatrix}1&1\0 & 1end{pmatrix}^ell$ lies in $Gamma$?
group-theory modular-group
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
$endgroup$
Let $G$ be a group, not necessarily finite. If $H$ is a normal subgroup of $G$ of a finite index, say $(G:H)=n$, then for every $gin G$ we have $g^nin H$. Does this statement remain valid if do not assume $H$ to be normal?
In particular let $SL_2(mathbb Z)$ be the modular group, and let $Gammasubset SL_2(mathbb Z)$ be a subgroup of a finite index. Does there exists a positive integer $ell$ such that $begin{pmatrix}1&1\0 & 1end{pmatrix}^ell$ lies in $Gamma$?
group-theory modular-group
group-theory modular-group
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
asked Dec 18 '18 at 17:10
ShimrodShimrod
23229
23229
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26
add a comment |
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes.
The set ${ H, gH, g^2H, dots , g^nH }$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).
From $$g^aH=g^bH$$
it follows that $g^{a-b} in H$.
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
$endgroup$
add a comment |
$begingroup$
For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)
Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $rho : Gto mathfrak{S}G/H$.
Its kernel $K$ is contained in $H$ : indeed if $xin K$, then $H= rho(x)(H)= xH$, so $xin H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/Ksimeq mathrm{Im}rho subset mathfrak{S}G/Hsimeq mathfrak{S}_{|G/H|}$).
Thus if $xin G, x^nin K$ for some $n$, thus $x^nin H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
$endgroup$
$begingroup$
It's odd how the symbol/rendering of$mathfrak{S}$(i.e., $mathfrak{S}$) looks like a $G$.
$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
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%2f3045416%2fdoes-every-subgroup-of-finite-index-contain-a-power-of-each-element-of-the-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes.
The set ${ H, gH, g^2H, dots , g^nH }$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).
From $$g^aH=g^bH$$
it follows that $g^{a-b} in H$.
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
$endgroup$
add a comment |
$begingroup$
Yes.
The set ${ H, gH, g^2H, dots , g^nH }$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).
From $$g^aH=g^bH$$
it follows that $g^{a-b} in H$.
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
$endgroup$
add a comment |
$begingroup$
Yes.
The set ${ H, gH, g^2H, dots , g^nH }$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).
From $$g^aH=g^bH$$
it follows that $g^{a-b} in H$.
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
$endgroup$
Yes.
The set ${ H, gH, g^2H, dots , g^nH }$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).
From $$g^aH=g^bH$$
it follows that $g^{a-b} in H$.
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
answered Dec 18 '18 at 17:15
CrostulCrostul
28.2k22352
28.2k22352
add a comment |
add a comment |
$begingroup$
For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)
Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $rho : Gto mathfrak{S}G/H$.
Its kernel $K$ is contained in $H$ : indeed if $xin K$, then $H= rho(x)(H)= xH$, so $xin H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/Ksimeq mathrm{Im}rho subset mathfrak{S}G/Hsimeq mathfrak{S}_{|G/H|}$).
Thus if $xin G, x^nin K$ for some $n$, thus $x^nin H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
$endgroup$
$begingroup$
It's odd how the symbol/rendering of$mathfrak{S}$(i.e., $mathfrak{S}$) looks like a $G$.
$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
add a comment |
$begingroup$
For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)
Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $rho : Gto mathfrak{S}G/H$.
Its kernel $K$ is contained in $H$ : indeed if $xin K$, then $H= rho(x)(H)= xH$, so $xin H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/Ksimeq mathrm{Im}rho subset mathfrak{S}G/Hsimeq mathfrak{S}_{|G/H|}$).
Thus if $xin G, x^nin K$ for some $n$, thus $x^nin H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
$endgroup$
$begingroup$
It's odd how the symbol/rendering of$mathfrak{S}$(i.e., $mathfrak{S}$) looks like a $G$.
$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
add a comment |
$begingroup$
For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)
Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $rho : Gto mathfrak{S}G/H$.
Its kernel $K$ is contained in $H$ : indeed if $xin K$, then $H= rho(x)(H)= xH$, so $xin H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/Ksimeq mathrm{Im}rho subset mathfrak{S}G/Hsimeq mathfrak{S}_{|G/H|}$).
Thus if $xin G, x^nin K$ for some $n$, thus $x^nin H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
$endgroup$
For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)
Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $rho : Gto mathfrak{S}G/H$.
Its kernel $K$ is contained in $H$ : indeed if $xin K$, then $H= rho(x)(H)= xH$, so $xin H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/Ksimeq mathrm{Im}rho subset mathfrak{S}G/Hsimeq mathfrak{S}_{|G/H|}$).
Thus if $xin G, x^nin K$ for some $n$, thus $x^nin H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
answered Dec 18 '18 at 17:19
MaxMax
15.3k11143
15.3k11143
$begingroup$
It's odd how the symbol/rendering of$mathfrak{S}$(i.e., $mathfrak{S}$) looks like a $G$.
$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
add a comment |
$begingroup$
It's odd how the symbol/rendering of$mathfrak{S}$(i.e., $mathfrak{S}$) looks like a $G$.
$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
$begingroup$
It's odd how the symbol/rendering of
$mathfrak{S}$ (i.e., $mathfrak{S}$) looks like a $G$.$endgroup$
– Shaun
Dec 18 '18 at 17:29
$begingroup$
It's odd how the symbol/rendering of
$mathfrak{S}$ (i.e., $mathfrak{S}$) looks like a $G$.$endgroup$
– Shaun
Dec 18 '18 at 17:29
2
2
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
$begingroup$
@Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve
$endgroup$
– Max
Dec 18 '18 at 17:41
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%2f3045416%2fdoes-every-subgroup-of-finite-index-contain-a-power-of-each-element-of-the-group%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
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 18 '18 at 17:26