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Needed: Help with proving commutitivity

  1. Feb 14, 2008 #1
    Let (G,*) a group and a an element of G such that ax³a=x for all x element of G. Show that (G,*) is commutative. Note that * = multiplication. I am not sure where it is that i should be starting, any help would be appreciated. Thank You.
  2. jcsd
  3. Feb 14, 2008 #2
    What is a? Is it in G does it have an inverse?
  4. Feb 14, 2008 #3
    What is written is the exact question not short handed and that is all that is given.
  5. Feb 14, 2008 #4
    it says that a is an element of G, so a is in G but im not sure if it has an inverse. But since it says that G is a group under the operation that automatically means that any element in G has an inverse and that a neutral element exists within G.
  6. Feb 14, 2008 #5
    I can't figure it out. My excuse is I haven't taken a course on group theory. Here is my feeble attempt.

    [tex]x_1 x_2 = y [/tex]
    [tex]x_1 = y (x_2)^{-1}[/tex]
    [tex](x_1)^3 = x_1 y (x_2)^{-1} x_1[/tex]
    [tex]a (x_1)^3 a = a x_1 y (x_2)^{-1} x_1 a[/tex]
    [tex]a x_1 a = a x_1 y (x_2)^{-1} x_1 a[/tex]
    [tex]x_1=x_1 y (x_2)^{-1} x_1[/tex]
    [tex]I=y (x_2)^{-1} x_1[/tex]
    [tex] (x_1)^{-1} x_2 =y[/tex]
    Last edited: Feb 14, 2008
  7. Feb 14, 2008 #6
    I looked up the definition in wikipedia.


    Every element of a group has an inverse.
  8. Feb 15, 2008 #7


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    The following is the first method that occured to me. Can probably be shortened.
    Try substituting in x=1, so a2=1.
    Then substitute in ax in place of x. a(ax)3a=ax.
    Substitute ax3a in place of x on the rhs. a(ax)3a=aax3a
    Expand and cancel the aa on the left. xaxaxa=x3a.
    Cancel the x on the left and the xa on the right. axa = x.
    Substitute for x on the rhs in the original expression. ax3a=axa.
    Cancel ax on the left and a on the right. x2=1 (*).

    Now it's fairly standard. Substitute xy for x in (*). xyxy=1.
    Multiply on left by x and right by y. x2yxy2=xy.
    From (*) we know that x2=y2=1, so yx=xy.
  9. Feb 15, 2008 #8
    Since [itex]a\,x^3\,a=x,\, \forall x\in G[/itex] then for [itex]x=e[/itex] we have [itex]a\,e^3\,a=e\Rightarrow a^2=e\Rightarrow a=a^{-1}[/itex]. Thus

    [tex] a\,x^3\,a=x\Rightarrow a\,x^3=x\,a \quad \text{and}\quad x^3\,a=a\,x \quad(1)[/tex]

    which yields

    [tex]x^3\,a=a\,\,x^3\quad (2)[/tex]

    Then for [itex]x,\,y \in G[/itex] we have

    [tex]x\,y=x\,e\,y\,e\Rightarrow x\,y= x\,a^2\,y\,a^2\Rightarrow x\,y=(x\,a)\,a\,(y\,a)\,a\overset{(1)}\Rightarrow x\,y=(a\,x^3)\,a\,(a\,y^3)\,a\Rightarrow x\,y=a\,x^3\,y^3\,a\overset{(2)}\Rightarrow x\,y=x^3\,a\,a\,y^3\Rightarrow x\,y=x^3\,y^3\Rightarrow[/tex]

    [tex]x^{-1}\,x\,y\,y^{-1}=x^{-1}\,x^3\,y^3\,y^{-1}\Rightarrow e=x^2\,y^2\Rightarrow x^{-1}\,y^{-1}=x\,y\Rightarrow (y\,x)^{-1}=x\,y \quad \forall (x,y)\in G \quad (\ast)[/tex]​

    Now let in [itex](\ast)\, y=e\Rightarrow x^{-1}=x \quad \forall x\in G[/itex], thus

    [tex](\ast)\Rightarrow y\,x=x\,y \quad \forall (x,y) \in G[/tex]​
  10. Feb 15, 2008 #9
    Oupppsss! gel was faster! :smile:
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