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
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oldest
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$begingroup$
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
15.7k11143
$endgroup$
$begingroup$
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 |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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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
15.7k11143
$endgroup$
$begingroup$
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
15.7k11143
$endgroup$
$begingroup$
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
15.7k11143
$endgroup$
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
15.7k11143
answered Dec 18 '18 at 17:19
MaxMax
15.7k11143
answered Dec 18 '18 at 17:19
MaxMax
15.7k11143
answered Dec 18 '18 at 17:19
MaxMax
15.7k11143
15.7k11143
$begingroup$
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 |
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$begingroup$
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