Subgroups of the group of motions on the plane fixing a point is the conjugate of the group of all orthogonal...












0












$begingroup$


How do I prove that subgroups $O'$ of the group of motions on the plane fixing a point $p$ (say) is the conjugate of the group of all orthogonal transformations by translations i.e. $O' = t_{p} O {t_{p}}^{-1}$ where $O$ is the group of all orthogonal transformations?



It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'$. How do I prove the other way round? Please help me in this regard.



Thank you very much.






abstract-algebra group-theory rigid-transformation







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$endgroup$












  • $begingroup$
    Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 16:42










  • $begingroup$
    I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 16:57












  • $begingroup$
    As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 17:15












  • $begingroup$
    the semidirect product bit is inessential (although helpful)
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 17:15


















0












$begingroup$


How do I prove that subgroups $O'$ of the group of motions on the plane fixing a point $p$ (say) is the conjugate of the group of all orthogonal transformations by translations i.e. $O' = t_{p} O {t_{p}}^{-1}$ where $O$ is the group of all orthogonal transformations?



It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'$. How do I prove the other way round? Please help me in this regard.



Thank you very much.






abstract-algebra group-theory rigid-transformation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 16:42










  • $begingroup$
    I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 16:57












  • $begingroup$
    As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 17:15












  • $begingroup$
    the semidirect product bit is inessential (although helpful)
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 17:15
















0












0








0





$begingroup$


How do I prove that subgroups $O'$ of the group of motions on the plane fixing a point $p$ (say) is the conjugate of the group of all orthogonal transformations by translations i.e. $O' = t_{p} O {t_{p}}^{-1}$ where $O$ is the group of all orthogonal transformations?



It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'$. How do I prove the other way round? Please help me in this regard.



Thank you very much.






abstract-algebra group-theory rigid-transformation







share|cite|improve this question









$endgroup$




How do I prove that subgroups $O'$ of the group of motions on the plane fixing a point $p$ (say) is the conjugate of the group of all orthogonal transformations by translations i.e. $O' = t_{p} O {t_{p}}^{-1}$ where $O$ is the group of all orthogonal transformations?



It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'$. How do I prove the other way round? Please help me in this regard.



Thank you very much.






abstract-algebra group-theory rigid-transformation




abstract-algebra group-theory rigid-transformation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 16:31









Dbchatto67Dbchatto67

1,728219




1,728219












  • $begingroup$
    Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 16:42










  • $begingroup$
    I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 16:57












  • $begingroup$
    As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 17:15












  • $begingroup$
    the semidirect product bit is inessential (although helpful)
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 17:15




















  • $begingroup$
    Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 16:42










  • $begingroup$
    I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 16:57












  • $begingroup$
    As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
    $endgroup$
    – Dbchatto67
    Dec 24 '18 at 17:15












  • $begingroup$
    the semidirect product bit is inessential (although helpful)
    $endgroup$
    – Andres Mejia
    Dec 24 '18 at 17:15


















$begingroup$
Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
$endgroup$
– Andres Mejia
Dec 24 '18 at 16:42




$begingroup$
Show that the group of translations $mathbb R^2$ is normal. Can you show that $mathrm{Isom}(mathbb R^2)/mathbb R^2) cong O(2)$? And in fact, there is a section $O(2) to mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down.
$endgroup$
– Andres Mejia
Dec 24 '18 at 16:42












$begingroup$
I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:57






$begingroup$
I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product.
$endgroup$
– Dbchatto67
Dec 24 '18 at 16:57














$begingroup$
As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
$endgroup$
– Dbchatto67
Dec 24 '18 at 17:15






$begingroup$
As far as I know $G/Bbb R^2 simeq O(2)$ where $G$ is the group of motoins in the plane.
$endgroup$
– Dbchatto67
Dec 24 '18 at 17:15














$begingroup$
the semidirect product bit is inessential (although helpful)
$endgroup$
– Andres Mejia
Dec 24 '18 at 17:15






$begingroup$
the semidirect product bit is inessential (although helpful)
$endgroup$
– Andres Mejia
Dec 24 '18 at 17:15












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$begingroup$

We take an element $m' in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m in O.$ Therefore $m' in t_{p} O {t_{p}}^{-1}.$ So $O' subseteq t_{p} O {t_{p}}^{-1}.$ Also It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'.$ Hence $t_{p} O {t_{p}}^{-1} = O'.$ This completes the proof.



QED






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    $begingroup$

    We take an element $m' in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m in O.$ Therefore $m' in t_{p} O {t_{p}}^{-1}.$ So $O' subseteq t_{p} O {t_{p}}^{-1}.$ Also It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'.$ Hence $t_{p} O {t_{p}}^{-1} = O'.$ This completes the proof.



    QED






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We take an element $m' in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m in O.$ Therefore $m' in t_{p} O {t_{p}}^{-1}.$ So $O' subseteq t_{p} O {t_{p}}^{-1}.$ Also It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'.$ Hence $t_{p} O {t_{p}}^{-1} = O'.$ This completes the proof.



      QED






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We take an element $m' in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m in O.$ Therefore $m' in t_{p} O {t_{p}}^{-1}.$ So $O' subseteq t_{p} O {t_{p}}^{-1}.$ Also It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'.$ Hence $t_{p} O {t_{p}}^{-1} = O'.$ This completes the proof.



        QED






        share|cite|improve this answer









        $endgroup$



        We take an element $m' in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m in O.$ Therefore $m' in t_{p} O {t_{p}}^{-1}.$ So $O' subseteq t_{p} O {t_{p}}^{-1}.$ Also It is easy to see that $t_{p} O {t_{p}}^{-1} subseteq O'.$ Hence $t_{p} O {t_{p}}^{-1} = O'.$ This completes the proof.



        QED







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        answered Jan 1 at 9:53









        Dbchatto67Dbchatto67

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