Find a group $G$ that has elements $V$ and $H$ both with order $2$ and the order of $VH$ is $3$.












1















Find a group $G$ that has elements $V$ and $H$ both with order $2$ and the order of $VH$ is $3$.




This question appeared on my final and even though it looks like an easy question I had difficulty finding a group.






group-theory examples-counterexamples







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  • A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
    – Shaun
    Nov 28 '18 at 23:32
















1















Find a group $G$ that has elements $V$ and $H$ both with order $2$ and the order of $VH$ is $3$.




This question appeared on my final and even though it looks like an easy question I had difficulty finding a group.






group-theory examples-counterexamples







share|cite|improve this question
























  • A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
    – Shaun
    Nov 28 '18 at 23:32














1












1








1








Find a group $G$ that has elements $V$ and $H$ both with order $2$ and the order of $VH$ is $3$.




This question appeared on my final and even though it looks like an easy question I had difficulty finding a group.






group-theory examples-counterexamples







share|cite|improve this question
















Find a group $G$ that has elements $V$ and $H$ both with order $2$ and the order of $VH$ is $3$.




This question appeared on my final and even though it looks like an easy question I had difficulty finding a group.






group-theory examples-counterexamples




group-theory examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 23:30









Shaun

8,820113681




8,820113681










asked Dec 18 '13 at 20:30









MathematicalAnomalyMathematicalAnomaly

651515




651515












  • A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
    – Shaun
    Nov 28 '18 at 23:32


















  • A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
    – Shaun
    Nov 28 '18 at 23:32
















A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
– Shaun
Nov 28 '18 at 23:32




A rather flippant answer would be to let $G$ be the group defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$
– Shaun
Nov 28 '18 at 23:32










3 Answers
3






active

oldest

votes


















5














$S_3$ gives such an example: Choose $V = (12)$ and $H = (23)$; then $VH$ is a $3$-cycle, and so has order $3$.




As a guide for finding this sort of example, it's immediate that the group cannot be abelian: This would mean that $VH$ has order $2$. The group order must be divisible by both $2$ and $3$, so $S_3$ is the smallest possible example. The next smallest non-abelian group satisfying these order conditions has order $12$.






share|cite|improve this answer





















  • Well now I feel stupid. I was only looking at Z D and U.
    – MathematicalAnomaly
    Dec 18 '13 at 20:36



















1














Consider the symmetries of an equilateral triangle ($D_3$ or $D_6$, depending on choice of notation). Reflecting across an axis is an operation of order two, but if we reflect over one axis and then a different axis, we get a rotation of the triangle.






share|cite|improve this answer





























    0














    Let $G$ be defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$ By construction, then, $G$ is of the required form.






    share|cite|improve this answer























    • The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
      – Shaun
      Nov 29 '18 at 2:20










    • NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
      – Shaun
      Nov 29 '18 at 3:45











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    $S_3$ gives such an example: Choose $V = (12)$ and $H = (23)$; then $VH$ is a $3$-cycle, and so has order $3$.




    As a guide for finding this sort of example, it's immediate that the group cannot be abelian: This would mean that $VH$ has order $2$. The group order must be divisible by both $2$ and $3$, so $S_3$ is the smallest possible example. The next smallest non-abelian group satisfying these order conditions has order $12$.






    share|cite|improve this answer





















    • Well now I feel stupid. I was only looking at Z D and U.
      – MathematicalAnomaly
      Dec 18 '13 at 20:36
















    5














    $S_3$ gives such an example: Choose $V = (12)$ and $H = (23)$; then $VH$ is a $3$-cycle, and so has order $3$.




    As a guide for finding this sort of example, it's immediate that the group cannot be abelian: This would mean that $VH$ has order $2$. The group order must be divisible by both $2$ and $3$, so $S_3$ is the smallest possible example. The next smallest non-abelian group satisfying these order conditions has order $12$.






    share|cite|improve this answer





















    • Well now I feel stupid. I was only looking at Z D and U.
      – MathematicalAnomaly
      Dec 18 '13 at 20:36














    5












    5








    5






    $S_3$ gives such an example: Choose $V = (12)$ and $H = (23)$; then $VH$ is a $3$-cycle, and so has order $3$.




    As a guide for finding this sort of example, it's immediate that the group cannot be abelian: This would mean that $VH$ has order $2$. The group order must be divisible by both $2$ and $3$, so $S_3$ is the smallest possible example. The next smallest non-abelian group satisfying these order conditions has order $12$.






    share|cite|improve this answer












    $S_3$ gives such an example: Choose $V = (12)$ and $H = (23)$; then $VH$ is a $3$-cycle, and so has order $3$.




    As a guide for finding this sort of example, it's immediate that the group cannot be abelian: This would mean that $VH$ has order $2$. The group order must be divisible by both $2$ and $3$, so $S_3$ is the smallest possible example. The next smallest non-abelian group satisfying these order conditions has order $12$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 18 '13 at 20:32







    user61527



















    • Well now I feel stupid. I was only looking at Z D and U.
      – MathematicalAnomaly
      Dec 18 '13 at 20:36


















    • Well now I feel stupid. I was only looking at Z D and U.
      – MathematicalAnomaly
      Dec 18 '13 at 20:36
















    Well now I feel stupid. I was only looking at Z D and U.
    – MathematicalAnomaly
    Dec 18 '13 at 20:36




    Well now I feel stupid. I was only looking at Z D and U.
    – MathematicalAnomaly
    Dec 18 '13 at 20:36











    1














    Consider the symmetries of an equilateral triangle ($D_3$ or $D_6$, depending on choice of notation). Reflecting across an axis is an operation of order two, but if we reflect over one axis and then a different axis, we get a rotation of the triangle.






    share|cite|improve this answer


























      1














      Consider the symmetries of an equilateral triangle ($D_3$ or $D_6$, depending on choice of notation). Reflecting across an axis is an operation of order two, but if we reflect over one axis and then a different axis, we get a rotation of the triangle.






      share|cite|improve this answer
























        1












        1








        1






        Consider the symmetries of an equilateral triangle ($D_3$ or $D_6$, depending on choice of notation). Reflecting across an axis is an operation of order two, but if we reflect over one axis and then a different axis, we get a rotation of the triangle.






        share|cite|improve this answer












        Consider the symmetries of an equilateral triangle ($D_3$ or $D_6$, depending on choice of notation). Reflecting across an axis is an operation of order two, but if we reflect over one axis and then a different axis, we get a rotation of the triangle.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '13 at 20:34









        AaronAaron

        15.8k22754




        15.8k22754























            0














            Let $G$ be defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$ By construction, then, $G$ is of the required form.






            share|cite|improve this answer























            • The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
              – Shaun
              Nov 29 '18 at 2:20










            • NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
              – Shaun
              Nov 29 '18 at 3:45
















            0














            Let $G$ be defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$ By construction, then, $G$ is of the required form.






            share|cite|improve this answer























            • The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
              – Shaun
              Nov 29 '18 at 2:20










            • NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
              – Shaun
              Nov 29 '18 at 3:45














            0












            0








            0






            Let $G$ be defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$ By construction, then, $G$ is of the required form.






            share|cite|improve this answer














            Let $G$ be defined by the group presentation $$langle V, Hmid V^2, H^2, (VH)^3rangle.$$ By construction, then, $G$ is of the required form.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 29 '18 at 2:11

























            answered Nov 29 '18 at 0:11









            ShaunShaun

            8,820113681




            8,820113681












            • The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
              – Shaun
              Nov 29 '18 at 2:20










            • NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
              – Shaun
              Nov 29 '18 at 3:45


















            • The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
              – Shaun
              Nov 29 '18 at 2:20










            • NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
              – Shaun
              Nov 29 '18 at 3:45
















            The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
            – Shaun
            Nov 29 '18 at 2:20




            The abelianisation of $G$ is given by $$begin{align}G^{ab}&=langle V, Hmid V^2, H^2, V^3H^3rangle^{ab}\ &=langle V, Hmid V^2, H^2, V=V^3=H^{-3}=H^{-1}rangle^{ab}\ &=langle Vmid V^2rangle\ &=Bbb Z_2,end{align}$$ which does not have an element of order $3$. Hence $G$, as defined in this answer, is not abelian.
            – Shaun
            Nov 29 '18 at 2:20












            NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
            – Shaun
            Nov 29 '18 at 3:45




            NB: The equalities in my previous comment should, in fact, be up to isomorphism only.
            – Shaun
            Nov 29 '18 at 3:45


















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