Symmetry group of equilateral triangle











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I have read at some places that the symmetry of equilateral triangle is C3v
as well as some places mention it to be D3.



The group tables for these two groups differ, hence they are not isomorphic.



Yet both these groups define symmetry of same shape.



Please, explain what is going on.










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    up vote
    1
    down vote

    favorite












    I have read at some places that the symmetry of equilateral triangle is C3v
    as well as some places mention it to be D3.



    The group tables for these two groups differ, hence they are not isomorphic.



    Yet both these groups define symmetry of same shape.



    Please, explain what is going on.










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have read at some places that the symmetry of equilateral triangle is C3v
      as well as some places mention it to be D3.



      The group tables for these two groups differ, hence they are not isomorphic.



      Yet both these groups define symmetry of same shape.



      Please, explain what is going on.










      share|cite|improve this question













      I have read at some places that the symmetry of equilateral triangle is C3v
      as well as some places mention it to be D3.



      The group tables for these two groups differ, hence they are not isomorphic.



      Yet both these groups define symmetry of same shape.



      Please, explain what is going on.







      group-theory finite-groups symmetric-groups






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      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 21 at 9:17









      Chetan Waghela

      637




      637






















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          The symmetry group of an equilateral triangle is the dihedral group $D_3$ with $6$ elements. It is a non-abelian group and hence isomorphic to $S_3$, since $C_6$ is abelian and there are only two different groups of order $6$. So there is one and only one symmetry group of the regular $3$-gon up to isomorphism. In particular, $C_{3v}cong D_3$.



          Reference: see page $105$ here.






          share|cite|improve this answer























          • I am not a mathematician what does the symbol in last sentence mean ?
            – Chetan Waghela
            Nov 21 at 9:51






          • 1




            $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
            – Dietrich Burde
            Nov 21 at 10:01













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          up vote
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          accepted










          The symmetry group of an equilateral triangle is the dihedral group $D_3$ with $6$ elements. It is a non-abelian group and hence isomorphic to $S_3$, since $C_6$ is abelian and there are only two different groups of order $6$. So there is one and only one symmetry group of the regular $3$-gon up to isomorphism. In particular, $C_{3v}cong D_3$.



          Reference: see page $105$ here.






          share|cite|improve this answer























          • I am not a mathematician what does the symbol in last sentence mean ?
            – Chetan Waghela
            Nov 21 at 9:51






          • 1




            $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
            – Dietrich Burde
            Nov 21 at 10:01

















          up vote
          1
          down vote



          accepted










          The symmetry group of an equilateral triangle is the dihedral group $D_3$ with $6$ elements. It is a non-abelian group and hence isomorphic to $S_3$, since $C_6$ is abelian and there are only two different groups of order $6$. So there is one and only one symmetry group of the regular $3$-gon up to isomorphism. In particular, $C_{3v}cong D_3$.



          Reference: see page $105$ here.






          share|cite|improve this answer























          • I am not a mathematician what does the symbol in last sentence mean ?
            – Chetan Waghela
            Nov 21 at 9:51






          • 1




            $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
            – Dietrich Burde
            Nov 21 at 10:01















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The symmetry group of an equilateral triangle is the dihedral group $D_3$ with $6$ elements. It is a non-abelian group and hence isomorphic to $S_3$, since $C_6$ is abelian and there are only two different groups of order $6$. So there is one and only one symmetry group of the regular $3$-gon up to isomorphism. In particular, $C_{3v}cong D_3$.



          Reference: see page $105$ here.






          share|cite|improve this answer














          The symmetry group of an equilateral triangle is the dihedral group $D_3$ with $6$ elements. It is a non-abelian group and hence isomorphic to $S_3$, since $C_6$ is abelian and there are only two different groups of order $6$. So there is one and only one symmetry group of the regular $3$-gon up to isomorphism. In particular, $C_{3v}cong D_3$.



          Reference: see page $105$ here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 9:40

























          answered Nov 21 at 9:35









          Dietrich Burde

          76.9k64286




          76.9k64286












          • I am not a mathematician what does the symbol in last sentence mean ?
            – Chetan Waghela
            Nov 21 at 9:51






          • 1




            $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
            – Dietrich Burde
            Nov 21 at 10:01




















          • I am not a mathematician what does the symbol in last sentence mean ?
            – Chetan Waghela
            Nov 21 at 9:51






          • 1




            $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
            – Dietrich Burde
            Nov 21 at 10:01


















          I am not a mathematician what does the symbol in last sentence mean ?
          – Chetan Waghela
          Nov 21 at 9:51




          I am not a mathematician what does the symbol in last sentence mean ?
          – Chetan Waghela
          Nov 21 at 9:51




          1




          1




          $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
          – Dietrich Burde
          Nov 21 at 10:01






          $Gcong H$ means that $G$ and $H$ are isomorphic as groups. So we may consider them as the same group (since you deal with irreducible representations of groups I assume that you are familiar with isomorphisms. It also seems that you never have accepted any answer:) ).
          – Dietrich Burde
          Nov 21 at 10:01




















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