# Needed: Help with proving commutitivity

• Devoo
In summary: DIn summary, the conversation discusses how to show that a group (G,*) with an element a such that ax³a=x for all x in G is commutative. It is noted that a must exist in G and have an inverse, and it is shown through substitution and manipulation that x3a=a for all x in G. Using this, it is then proven that (G,*) is commutative by showing that xy=yx for all x and y in G.
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.

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

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

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 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:
Devoo said:
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.

Devoo said:
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.

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

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

Oupppsss! gel was faster!

## What is commutativity?

Commutativity is a mathematical property where the order of operations does not affect the final result. In other words, it does not matter which order you perform a set of operations, the end result will be the same.

## Why is proving commutativity important?

Proving commutativity is important because it helps us understand and solve complex mathematical problems. It also allows us to make generalizations and simplifications in mathematical equations.

## How do you prove commutativity?

There are various methods for proving commutativity, depending on the type of mathematical operation involved. Generally, it involves showing that the order of operations can be changed without affecting the final result. This can be done through algebraic manipulation, logical reasoning, or using specific mathematical properties.

## What are some examples of commutative operations?

Addition and multiplication are two common examples of commutative operations. For example, 2+3=3+2 and 5x4=4x5. Other examples include exponentiation, matrix addition, and scalar multiplication.

## Are all operations commutative?

No, not all operations are commutative. Subtraction and division, for example, are not commutative as changing the order of the numbers will result in a different outcome. In general, operations that involve subtraction, division, or non-commutative matrices are not commutative.

• Linear and Abstract Algebra
Replies
1
Views
695
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
801
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
16
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
917