why $x^k in x langle y rangle$ which does not contain $e$.?

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A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



i have taken from



A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



as i did n't undersatnds the enter image description here red line not getting in my head



pliz help me,,,,,,




abstract-algebra




share|cite|improve this question























    up vote
    -1
    down vote

    favorite












    A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



    i have taken from



    A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



    as i did n't undersatnds the enter image description here red line not getting in my head



    pliz help me,,,,,,




    abstract-algebra




    share|cite|improve this question





















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



      i have taken from



      A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



      as i did n't undersatnds the enter image description here red line not getting in my head



      pliz help me,,,,,,




      abstract-algebra




      share|cite|improve this question











      A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



      i have taken from



      A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $



      as i did n't undersatnds the enter image description here red line not getting in my head



      pliz help me,,,,,,





      abstract-algebra





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 28 at 20:57









      Messi fifa

      1648




      1648




















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          You have $x^2=y^-2$, hence $x^2$ is generated by $y$ and hence $x^2 in langle y rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 in xlangle y rangle$. But we have a problem here because the identity does not belong to any left coset of $langle y rangle$ which is not $langle y rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.






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          • @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
            – Messi fifa
            Jul 28 at 21:19






          • 1




            Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
            – Mark
            Jul 28 at 21:24










          • thanks u @Mark.
            – Messi fifa
            Jul 28 at 21:28










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          You have $x^2=y^-2$, hence $x^2$ is generated by $y$ and hence $x^2 in langle y rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 in xlangle y rangle$. But we have a problem here because the identity does not belong to any left coset of $langle y rangle$ which is not $langle y rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.






          share|cite|improve this answer





















          • @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
            – Messi fifa
            Jul 28 at 21:19






          • 1




            Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
            – Mark
            Jul 28 at 21:24










          • thanks u @Mark.
            – Messi fifa
            Jul 28 at 21:28














          up vote
          1
          down vote



          accepted










          You have $x^2=y^-2$, hence $x^2$ is generated by $y$ and hence $x^2 in langle y rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 in xlangle y rangle$. But we have a problem here because the identity does not belong to any left coset of $langle y rangle$ which is not $langle y rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.






          share|cite|improve this answer





















          • @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
            – Messi fifa
            Jul 28 at 21:19






          • 1




            Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
            – Mark
            Jul 28 at 21:24










          • thanks u @Mark.
            – Messi fifa
            Jul 28 at 21:28












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You have $x^2=y^-2$, hence $x^2$ is generated by $y$ and hence $x^2 in langle y rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 in xlangle y rangle$. But we have a problem here because the identity does not belong to any left coset of $langle y rangle$ which is not $langle y rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.






          share|cite|improve this answer













          You have $x^2=y^-2$, hence $x^2$ is generated by $y$ and hence $x^2 in langle y rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 in xlangle y rangle$. But we have a problem here because the identity does not belong to any left coset of $langle y rangle$ which is not $langle y rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 28 at 21:13









          Mark

          64919




          64919











          • @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
            – Messi fifa
            Jul 28 at 21:19






          • 1




            Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
            – Mark
            Jul 28 at 21:24










          • thanks u @Mark.
            – Messi fifa
            Jul 28 at 21:28
















          • @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
            – Messi fifa
            Jul 28 at 21:19






          • 1




            Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
            – Mark
            Jul 28 at 21:24










          • thanks u @Mark.
            – Messi fifa
            Jul 28 at 21:28















          @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
          – Messi fifa
          Jul 28 at 21:19




          @ Mark.....one more doubts why identity doesnot belong to any left coset of <y> ??..
          – Messi fifa
          Jul 28 at 21:19




          1




          1




          Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
          – Mark
          Jul 28 at 21:24




          Let's assume it does belong to $xlangle y rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^-k=x$ and hence $x in langle y rangle$ which is a contradiction.
          – Mark
          Jul 28 at 21:24












          thanks u @Mark.
          – Messi fifa
          Jul 28 at 21:28




          thanks u @Mark.
          – Messi fifa
          Jul 28 at 21:28












           

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