Needed: Help with proving commutitivity

[Devoo]

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.

 

[John Creighto]

What is a? Is it in G does it have an inverse?

 

[Devoo]

What is written is the exact question not short handed and that is all that is given.

 

[Devoo]

it says that a is an element of G, so a is in G but I am 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.

 

[John Creighto]

I can't figure it out. My excuse is I haven't taken a course on group theory. Here is my feeble attempt.

x_1 x_2 = y
x_1 = y (x_2)^{-1}
(x_1)^3 = x_1 y (x_2)^{-1} x_1
a (x_1)^3 a = a x_1 y (x_2)^{-1} x_1 a
a x_1 a = a x_1 y (x_2)^{-1} x_1 a
x_1=x_1 y (x_2)^{-1} x_1
I=y (x_2)^{-1} x_1
(x_1)^{-1} x_2 =y

 

[John Creighto]

it says that a is an element of G, so a is in G but I am 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.

I looked up the definition in wikipedia.

http://en.wikipedia.org/wiki/Group_theory

Every element of a group has an inverse.

 

[gel]

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.

The following is the first method that occurred 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.

 

[Rainbow Child]

Since a\,x^3\,a=x,\, \forall x\in G then for x=e we have a\,e^3\,a=e\Rightarrow a^2=e\Rightarrow a=a^{-1}. Thus

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

which yields

x^3\,a=a\,\,x^3\quad (2)

Then for x,\,y \in G we have

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

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)​

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

(\ast)\Rightarrow y\,x=x\,y \quad \forall (x,y) \in G​

 

[Rainbow Child]

Oupppsss! gel was faster!