Does every subgroup of finite index contain a power of each element of the group?












6












$begingroup$


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







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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
















6












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














6












6








6


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















9












$begingroup$

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 $g^{a-b} in H$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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 $n=[G:K]$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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












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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

    oldest

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    9












    $begingroup$

    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 $g^{a-b} in H$.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      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 $g^{a-b} in H$.






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        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 $g^{a-b} in H$.






        share|cite|improve this answer









        $endgroup$



        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 $g^{a-b} in H$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 17:15









        CrostulCrostul

        28.2k22352




        28.2k22352























            3












            $begingroup$

            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 $n=[G:K]$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            3












            $begingroup$

            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 $n=[G:K]$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














            3












            3








            3





            $begingroup$

            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 $n=[G:K]$.






            share|cite|improve this answer









            $endgroup$



            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 $n=[G:K]$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $begingroup$
              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


















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