If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that...












1












$begingroup$



If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that $aNa^{-1}= N$ for all $a$ in $G$. Can you find an upper bound for $[G:N]$?




Let, $a_1H,a_2H,..,a_mH$ is the complete list of left cosets of $H$ in $G$.
Then, $cup_{i=1}^{m} a_i H=G$
$implies cup_{i=1}^{m} a_i H a_j^{-1}=Gimplies cup_{i=1}^{m} b_i(a_j H a_j^{-1})=G$



for each $j=1,2,..m$ and $b_iin G$ such that, $b_ia_j=a_i$ for $i=1,2,..,m$.



Now, we consider the subgroup,
$N=cap_{xin G} xHx^{-1}=cap_{i=1}^{m} a_iHa_i^{-1}$.
Then, clearly $N$ is a normal subgroup of $G$ which is contained in $H$.
Since, each of $a_iHa_i^{-1}$ is of finite index in $G$.
Then, $cap a_iHa_i^{-1}$ has also finite index in $G$.
And we are done.
Also, it's easy to see that,
$[G:N]le[G:a_1Ha_1^{-1}][G:a_2Ha_2^{-1}]..........[G:a_mHa_m^{-1}]$



Can anyone check it?






group-theory proof-verification







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$endgroup$












  • $begingroup$
    There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 18:08
















1












$begingroup$



If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that $aNa^{-1}= N$ for all $a$ in $G$. Can you find an upper bound for $[G:N]$?




Let, $a_1H,a_2H,..,a_mH$ is the complete list of left cosets of $H$ in $G$.
Then, $cup_{i=1}^{m} a_i H=G$
$implies cup_{i=1}^{m} a_i H a_j^{-1}=Gimplies cup_{i=1}^{m} b_i(a_j H a_j^{-1})=G$



for each $j=1,2,..m$ and $b_iin G$ such that, $b_ia_j=a_i$ for $i=1,2,..,m$.



Now, we consider the subgroup,
$N=cap_{xin G} xHx^{-1}=cap_{i=1}^{m} a_iHa_i^{-1}$.
Then, clearly $N$ is a normal subgroup of $G$ which is contained in $H$.
Since, each of $a_iHa_i^{-1}$ is of finite index in $G$.
Then, $cap a_iHa_i^{-1}$ has also finite index in $G$.
And we are done.
Also, it's easy to see that,
$[G:N]le[G:a_1Ha_1^{-1}][G:a_2Ha_2^{-1}]..........[G:a_mHa_m^{-1}]$



Can anyone check it?






group-theory proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 18:08














1












1








1





$begingroup$



If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that $aNa^{-1}= N$ for all $a$ in $G$. Can you find an upper bound for $[G:N]$?




Let, $a_1H,a_2H,..,a_mH$ is the complete list of left cosets of $H$ in $G$.
Then, $cup_{i=1}^{m} a_i H=G$
$implies cup_{i=1}^{m} a_i H a_j^{-1}=Gimplies cup_{i=1}^{m} b_i(a_j H a_j^{-1})=G$



for each $j=1,2,..m$ and $b_iin G$ such that, $b_ia_j=a_i$ for $i=1,2,..,m$.



Now, we consider the subgroup,
$N=cap_{xin G} xHx^{-1}=cap_{i=1}^{m} a_iHa_i^{-1}$.
Then, clearly $N$ is a normal subgroup of $G$ which is contained in $H$.
Since, each of $a_iHa_i^{-1}$ is of finite index in $G$.
Then, $cap a_iHa_i^{-1}$ has also finite index in $G$.
And we are done.
Also, it's easy to see that,
$[G:N]le[G:a_1Ha_1^{-1}][G:a_2Ha_2^{-1}]..........[G:a_mHa_m^{-1}]$



Can anyone check it?






group-theory proof-verification







share|cite|improve this question









$endgroup$





If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that $aNa^{-1}= N$ for all $a$ in $G$. Can you find an upper bound for $[G:N]$?




Let, $a_1H,a_2H,..,a_mH$ is the complete list of left cosets of $H$ in $G$.
Then, $cup_{i=1}^{m} a_i H=G$
$implies cup_{i=1}^{m} a_i H a_j^{-1}=Gimplies cup_{i=1}^{m} b_i(a_j H a_j^{-1})=G$



for each $j=1,2,..m$ and $b_iin G$ such that, $b_ia_j=a_i$ for $i=1,2,..,m$.



Now, we consider the subgroup,
$N=cap_{xin G} xHx^{-1}=cap_{i=1}^{m} a_iHa_i^{-1}$.
Then, clearly $N$ is a normal subgroup of $G$ which is contained in $H$.
Since, each of $a_iHa_i^{-1}$ is of finite index in $G$.
Then, $cap a_iHa_i^{-1}$ has also finite index in $G$.
And we are done.
Also, it's easy to see that,
$[G:N]le[G:a_1Ha_1^{-1}][G:a_2Ha_2^{-1}]..........[G:a_mHa_m^{-1}]$



Can anyone check it?






group-theory proof-verification




group-theory proof-verification






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asked Dec 10 '18 at 17:28









Tom.Tom.

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15619












  • $begingroup$
    There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 18:08


















  • $begingroup$
    There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 18:08
















$begingroup$
There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
$endgroup$
– Matt Samuel
Dec 10 '18 at 18:08




$begingroup$
There's at least one hole. The intersection of two finite index subgroups could be empty, for example, yielding a subgroup that is not of finite index.
$endgroup$
– Matt Samuel
Dec 10 '18 at 18:08










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$begingroup$

You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.



Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.






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    $begingroup$

    You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.



    Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.



      Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.



        Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.






        share|cite|improve this answer











        $endgroup$



        You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.



        Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 0:51

























        answered Dec 10 '18 at 18:39









        C MonsourC Monsour

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        6,2191325






























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