Subgroup and conjugacy classFinite group with conjugacy class of order 2 has nontrivial normal subgroup?Groups with uniform bound on the order of conjugacy classesIf a normal subgroup shares elements with a conjugacy class, then it contains it entirely?Subgroup generated by conjugacy class normal?Normal Group and Conjugacy ClassExistence of g whose conjugacy class is disjoint from a proper subgroupSize of a Conjugacy classFinite groups with only one conjugacy class of maximal subgroupsComplement of a normal subgroup is single conjugacy classSize of conjugacy class in subgroup compared to size of conjugacy class in group

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Subgroup and conjugacy class


Finite group with conjugacy class of order 2 has nontrivial normal subgroup?Groups with uniform bound on the order of conjugacy classesIf a normal subgroup shares elements with a conjugacy class, then it contains it entirely?Subgroup generated by conjugacy class normal?Normal Group and Conjugacy ClassExistence of g whose conjugacy class is disjoint from a proper subgroupSize of a Conjugacy classFinite groups with only one conjugacy class of maximal subgroupsComplement of a normal subgroup is single conjugacy classSize of conjugacy class in subgroup compared to size of conjugacy class in group













3












$begingroup$


let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $exists Hleq G$ with H being a subgroup of G, that $|G/H|=n$



It easy to proof that this happens with G being a finite group but my problem starts when G is a infinite group because I cannot apply Lagrange theorem and because of that I don't have any ideia of who is H



Any hints?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What does Lagrange's theorem have to do with this problem?
    $endgroup$
    – the_fox
    May 6 at 1:33















3












$begingroup$


let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $exists Hleq G$ with H being a subgroup of G, that $|G/H|=n$



It easy to proof that this happens with G being a finite group but my problem starts when G is a infinite group because I cannot apply Lagrange theorem and because of that I don't have any ideia of who is H



Any hints?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What does Lagrange's theorem have to do with this problem?
    $endgroup$
    – the_fox
    May 6 at 1:33













3












3








3


3



$begingroup$


let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $exists Hleq G$ with H being a subgroup of G, that $|G/H|=n$



It easy to proof that this happens with G being a finite group but my problem starts when G is a infinite group because I cannot apply Lagrange theorem and because of that I don't have any ideia of who is H



Any hints?










share|cite|improve this question











$endgroup$




let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $exists Hleq G$ with H being a subgroup of G, that $|G/H|=n$



It easy to proof that this happens with G being a finite group but my problem starts when G is a infinite group because I cannot apply Lagrange theorem and because of that I don't have any ideia of who is H



Any hints?







abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 6 at 1:27









Robert Lewis

49.8k23268




49.8k23268










asked May 6 at 1:20









Enzo MassakiEnzo Massaki

184




184







  • 1




    $begingroup$
    What does Lagrange's theorem have to do with this problem?
    $endgroup$
    – the_fox
    May 6 at 1:33












  • 1




    $begingroup$
    What does Lagrange's theorem have to do with this problem?
    $endgroup$
    – the_fox
    May 6 at 1:33







1




1




$begingroup$
What does Lagrange's theorem have to do with this problem?
$endgroup$
– the_fox
May 6 at 1:33




$begingroup$
What does Lagrange's theorem have to do with this problem?
$endgroup$
– the_fox
May 6 at 1:33










2 Answers
2






active

oldest

votes


















3












$begingroup$

Consider



$F_x = f in G, ; fxf^-1 = x ; tag 1$



that is, $F_x subset G$ is the set of group elements which fix $x in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for



$a, b in F_x tag 2$



we have



$(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$



and the identity $e in G$ is clearly in $F_x$:



$exe^-1 = exe = xe = x; tag 4$



and



$a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$



Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have



$(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$



which shows that elements $gf in gF_x$ all take $x$ to $gxg^-1$ under conjugation; in fact, the conjugate $gxg^-1$ only depends on the coset $gF_x$ and not upon its representative $g$; for if



$g_1F_x = g_2F_x, tag 7$



then



$g_1 = g_1e = g_2 f_1 tag 8$



for some $f_1 in F_x$, whence



$g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$



it follows then, that there is a well-defined map



$phi: G/F_x to C_x; tag10$



$phi$ is injective, for if



$phi(g_1F_x) = phi(g_2F_x), tag11$



then



$g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
$Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$



$phi$ is also surjective, for



$phi(gF_x) = gxg^-1 tag13$



for any conjugate of $x$. Since $phi$ is a bijection, we conclude that



$vert G/F_x vert = vert C_x vert = n, tag14$



$OEDelta$.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Hint: Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements). Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      3












      $begingroup$

      Consider



      $F_x = f in G, ; fxf^-1 = x ; tag 1$



      that is, $F_x subset G$ is the set of group elements which fix $x in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for



      $a, b in F_x tag 2$



      we have



      $(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$



      and the identity $e in G$ is clearly in $F_x$:



      $exe^-1 = exe = xe = x; tag 4$



      and



      $a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$



      Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have



      $(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$



      which shows that elements $gf in gF_x$ all take $x$ to $gxg^-1$ under conjugation; in fact, the conjugate $gxg^-1$ only depends on the coset $gF_x$ and not upon its representative $g$; for if



      $g_1F_x = g_2F_x, tag 7$



      then



      $g_1 = g_1e = g_2 f_1 tag 8$



      for some $f_1 in F_x$, whence



      $g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$



      it follows then, that there is a well-defined map



      $phi: G/F_x to C_x; tag10$



      $phi$ is injective, for if



      $phi(g_1F_x) = phi(g_2F_x), tag11$



      then



      $g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
      $Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$



      $phi$ is also surjective, for



      $phi(gF_x) = gxg^-1 tag13$



      for any conjugate of $x$. Since $phi$ is a bijection, we conclude that



      $vert G/F_x vert = vert C_x vert = n, tag14$



      $OEDelta$.






      share|cite|improve this answer









      $endgroup$

















        3












        $begingroup$

        Consider



        $F_x = f in G, ; fxf^-1 = x ; tag 1$



        that is, $F_x subset G$ is the set of group elements which fix $x in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for



        $a, b in F_x tag 2$



        we have



        $(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$



        and the identity $e in G$ is clearly in $F_x$:



        $exe^-1 = exe = xe = x; tag 4$



        and



        $a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$



        Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have



        $(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$



        which shows that elements $gf in gF_x$ all take $x$ to $gxg^-1$ under conjugation; in fact, the conjugate $gxg^-1$ only depends on the coset $gF_x$ and not upon its representative $g$; for if



        $g_1F_x = g_2F_x, tag 7$



        then



        $g_1 = g_1e = g_2 f_1 tag 8$



        for some $f_1 in F_x$, whence



        $g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$



        it follows then, that there is a well-defined map



        $phi: G/F_x to C_x; tag10$



        $phi$ is injective, for if



        $phi(g_1F_x) = phi(g_2F_x), tag11$



        then



        $g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
        $Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$



        $phi$ is also surjective, for



        $phi(gF_x) = gxg^-1 tag13$



        for any conjugate of $x$. Since $phi$ is a bijection, we conclude that



        $vert G/F_x vert = vert C_x vert = n, tag14$



        $OEDelta$.






        share|cite|improve this answer









        $endgroup$















          3












          3








          3





          $begingroup$

          Consider



          $F_x = f in G, ; fxf^-1 = x ; tag 1$



          that is, $F_x subset G$ is the set of group elements which fix $x in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for



          $a, b in F_x tag 2$



          we have



          $(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$



          and the identity $e in G$ is clearly in $F_x$:



          $exe^-1 = exe = xe = x; tag 4$



          and



          $a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$



          Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have



          $(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$



          which shows that elements $gf in gF_x$ all take $x$ to $gxg^-1$ under conjugation; in fact, the conjugate $gxg^-1$ only depends on the coset $gF_x$ and not upon its representative $g$; for if



          $g_1F_x = g_2F_x, tag 7$



          then



          $g_1 = g_1e = g_2 f_1 tag 8$



          for some $f_1 in F_x$, whence



          $g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$



          it follows then, that there is a well-defined map



          $phi: G/F_x to C_x; tag10$



          $phi$ is injective, for if



          $phi(g_1F_x) = phi(g_2F_x), tag11$



          then



          $g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
          $Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$



          $phi$ is also surjective, for



          $phi(gF_x) = gxg^-1 tag13$



          for any conjugate of $x$. Since $phi$ is a bijection, we conclude that



          $vert G/F_x vert = vert C_x vert = n, tag14$



          $OEDelta$.






          share|cite|improve this answer









          $endgroup$



          Consider



          $F_x = f in G, ; fxf^-1 = x ; tag 1$



          that is, $F_x subset G$ is the set of group elements which fix $x in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for



          $a, b in F_x tag 2$



          we have



          $(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$



          and the identity $e in G$ is clearly in $F_x$:



          $exe^-1 = exe = xe = x; tag 4$



          and



          $a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$



          Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have



          $(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$



          which shows that elements $gf in gF_x$ all take $x$ to $gxg^-1$ under conjugation; in fact, the conjugate $gxg^-1$ only depends on the coset $gF_x$ and not upon its representative $g$; for if



          $g_1F_x = g_2F_x, tag 7$



          then



          $g_1 = g_1e = g_2 f_1 tag 8$



          for some $f_1 in F_x$, whence



          $g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$



          it follows then, that there is a well-defined map



          $phi: G/F_x to C_x; tag10$



          $phi$ is injective, for if



          $phi(g_1F_x) = phi(g_2F_x), tag11$



          then



          $g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
          $Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$



          $phi$ is also surjective, for



          $phi(gF_x) = gxg^-1 tag13$



          for any conjugate of $x$. Since $phi$ is a bijection, we conclude that



          $vert G/F_x vert = vert C_x vert = n, tag14$



          $OEDelta$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 6 at 4:59









          Robert LewisRobert Lewis

          49.8k23268




          49.8k23268





















              2












              $begingroup$

              Hint: Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements). Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Hint: Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements). Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Hint: Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements). Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements). Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 6 at 1:37









                  Eric TowersEric Towers

                  34.3k22371




                  34.3k22371



























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                      Cegueira Índice Epidemioloxía | Deficiencia visual | Tipos de cegueira | Principais causas de cegueira | Tratamento | Técnicas de adaptación e axudas | Vida dos cegos | Primeiros auxilios | Crenzas respecto das persoas cegas | Crenzas das persoas cegas | O neno deficiente visual | Aspectos psicolóxicos da cegueira | Notas | Véxase tamén | Menú de navegación54.054.154.436928256blindnessDicionario da Real Academia GalegaPortal das Palabras"International Standards: Visual Standards — Aspects and Ranges of Vision Loss with Emphasis on Population Surveys.""Visual impairment and blindness""Presentan un plan para previr a cegueira"o orixinalACCDV Associació Catalana de Cecs i Disminuïts Visuals - PMFTrachoma"Effect of gene therapy on visual function in Leber's congenital amaurosis"1844137110.1056/NEJMoa0802268Cans guía - os mellores amigos dos cegosArquivadoEscola de cans guía para cegos en Mortágua, PortugalArquivado"Tecnología para ciegos y deficientes visuales. Recopilación de recursos gratuitos en la Red""Colorino""‘COL.diesis’, escuchar los sonidos del color""COL.diesis: Transforming Colour into Melody and Implementing the Result in a Colour Sensor Device"o orixinal"Sistema de desarrollo de sinestesia color-sonido para invidentes utilizando un protocolo de audio""Enseñanza táctil - geometría y color. Juegos didácticos para niños ciegos y videntes""Sistema Constanz"L'ocupació laboral dels cecs a l'Estat espanyol està pràcticament equiparada a la de les persones amb visió, entrevista amb Pedro ZuritaONCE (Organización Nacional de Cegos de España)Prevención da cegueiraDescrición de deficiencias visuais (Disc@pnet)Braillín, un boneco atractivo para calquera neno, con ou sen discapacidade, que permite familiarizarse co sistema de escritura e lectura brailleAxudas Técnicas36838ID00897494007150-90057129528256DOID:1432HP:0000618D001766C10.597.751.941.162C97109C0155020