# Needed: Help with proving commutitivity

1. Feb 14, 2008

### 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.

2. Feb 14, 2008

### John Creighto

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

3. Feb 14, 2008

### Devoo

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

4. Feb 14, 2008

### Devoo

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.

5. Feb 14, 2008

### 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.

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

Last edited: Feb 14, 2008
6. Feb 14, 2008

### John Creighto

I looked up the definition in wikipedia.

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

Every element of a group has an inverse.

7. Feb 15, 2008

### gel

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.

8. Feb 15, 2008

### 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

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

9. Feb 15, 2008

### Rainbow Child

Oupppsss! gel was faster!