# Thread: Abelian group

1. ## Abelian group

Hi, need a bit of help here.

The set of invertible affine transformations of R is a group under composition. Is this group Abelian? Prove it if it is, and if it is not, provide a counter example.

Not sure what the answer is.

Thank You for any help.

2. ## Re: Abelian group

Let $\displaystyle s(x)=x+1$ and $\displaystyle t(x)=2x$ over the field $\displaystyle \mathbb{R}$. Then $\displaystyle s,t$ are invertible affine transformations. But $\displaystyle s\circ t=2x+1\neq 2x+2=t\circ s$. So the group is not abelian.

EDIT: Oops, I see the transformation has to be over $\displaystyle \mathbb{R}$. Example has been changed to accommodate this.

3. ## Re: Abelian group

Originally Posted by shelford
Hi, need a bit of help here.

The set of invertible affine transformations of R is a group under composition. Is this group Abelian? Prove it if it is, and if it is not, provide a counter example.

Not sure what the answer is.

Thank You for any help.
Well an affine transformation is of the form

$\displaystyle T_1(x)=mx+b$ so create another one $\displaystyle T_2(x)=nx+c$

Now just compute

$\displaystyle T_1 \circ T_2$ and $\displaystyle T_2 \circ T_1$

If it is Abelian what must be true?

4. ## Re: Abelian group

I would like to point out a tangential consideration. This question has an interesting application in the field of data acquisition. I was in a situation once where I was taking in data on a computer, applying one slope and offset (affine transformation), and then applying another slope and offset. The purpose of applying these slopes and offsets is to convert the raw voltage data that you nearly always get off of your data acquisition hardware, into engineering units like psig. I needed to know whether I would get the same result if I switched the order of application. The answer to the question in this thread was also the answer to my query.

5. ## Re: Abelian group

Thanks to all for the great help. I understand it now.