Showing $S_{n}$ is not solvable for $ngeq 5$












1












$begingroup$


I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof



Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.



$(124)(142)=e$



$(135)(153)=e$



$(123)=(124)(135)(142)(153)$



Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$



$f[(124)]=x$



$f[(135)]=y$



$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$



Therefore $(123)in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.



I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.






abstract-algebra group-theory galois-theory symmetric-groups







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof



    Let $G$ be a group of permutations of five objects or more that
    include all cyclic permutations of three elements.



    $(124)(142)=e$



    $(135)(153)=e$



    $(123)=(124)(135)(142)(153)$



    Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$



    $f[(124)]=x$



    $f[(135)]=y$



    $f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$



    Therefore $(123)in H$. This is true for any cyclic permutation of three
    elements. Therefore, $G$ is not solvable.



    I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.






    abstract-algebra group-theory galois-theory symmetric-groups







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof



      Let $G$ be a group of permutations of five objects or more that
      include all cyclic permutations of three elements.



      $(124)(142)=e$



      $(135)(153)=e$



      $(123)=(124)(135)(142)(153)$



      Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$



      $f[(124)]=x$



      $f[(135)]=y$



      $f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$



      Therefore $(123)in H$. This is true for any cyclic permutation of three
      elements. Therefore, $G$ is not solvable.



      I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.






      abstract-algebra group-theory galois-theory symmetric-groups







      share|cite|improve this question











      $endgroup$




      I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof



      Let $G$ be a group of permutations of five objects or more that
      include all cyclic permutations of three elements.



      $(124)(142)=e$



      $(135)(153)=e$



      $(123)=(124)(135)(142)(153)$



      Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$



      $f[(124)]=x$



      $f[(135)]=y$



      $f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$



      Therefore $(123)in H$. This is true for any cyclic permutation of three
      elements. Therefore, $G$ is not solvable.



      I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.






      abstract-algebra group-theory galois-theory symmetric-groups




      abstract-algebra group-theory galois-theory symmetric-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 19:58







      Alberto Navarro

















      asked Dec 25 '18 at 19:52









      Alberto NavarroAlberto Navarro

      306




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          1 Answer
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          $begingroup$

          That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 1:28










          • $begingroup$
            By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 6:34












          • $begingroup$
            Yes, that's it.
            $endgroup$
            – José Carlos Santos
            Dec 26 '18 at 10:54











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          active

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          $begingroup$

          That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 1:28










          • $begingroup$
            By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 6:34












          • $begingroup$
            Yes, that's it.
            $endgroup$
            – José Carlos Santos
            Dec 26 '18 at 10:54
















          2












          $begingroup$

          That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 1:28










          • $begingroup$
            By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 6:34












          • $begingroup$
            Yes, that's it.
            $endgroup$
            – José Carlos Santos
            Dec 26 '18 at 10:54














          2












          2








          2





          $begingroup$

          That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.






          share|cite|improve this answer









          $endgroup$



          That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 19:59









          José Carlos SantosJosé Carlos Santos

          166k22132235




          166k22132235












          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 1:28










          • $begingroup$
            By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 6:34












          • $begingroup$
            Yes, that's it.
            $endgroup$
            – José Carlos Santos
            Dec 26 '18 at 10:54


















          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 1:28










          • $begingroup$
            By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
            $endgroup$
            – Alberto Navarro
            Dec 26 '18 at 6:34












          • $begingroup$
            Yes, that's it.
            $endgroup$
            – José Carlos Santos
            Dec 26 '18 at 10:54
















          $begingroup$
          Got it, thanks!
          $endgroup$
          – Alberto Navarro
          Dec 26 '18 at 1:28




          $begingroup$
          Got it, thanks!
          $endgroup$
          – Alberto Navarro
          Dec 26 '18 at 1:28












          $begingroup$
          By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
          $endgroup$
          – Alberto Navarro
          Dec 26 '18 at 6:34






          $begingroup$
          By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
          $endgroup$
          – Alberto Navarro
          Dec 26 '18 at 6:34














          $begingroup$
          Yes, that's it.
          $endgroup$
          – José Carlos Santos
          Dec 26 '18 at 10:54




          $begingroup$
          Yes, that's it.
          $endgroup$
          – José Carlos Santos
          Dec 26 '18 at 10:54


















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