Ytukyg

I'm trying to answer the following question from Herstein: Topics in Algebra (1st) section 2.5 Question 7-9.

Given the mapping:
$$T_{ab} : xrightarrow ax+b $$
Let $G = {T_{ab}| a not =0}$, then G is a group under composition of mappings, where
$$T_{ab} circ T_{cd} = T_{(ac)(ad+b)}$$

Let H be a subgroup of G, $ H ={T_{ab}in G|a text{is rational}}.$

The question is to list all right cosets of H in G, and show they are left cosets of H in G.

The right cosets, it seems to me, would be $$T_{ab} circ T_{rd} = T_{(ar)(ad+b)}$$

given $T_{rd} in G $ for r irrational.

The left cosets would be $$T_{rd} circ T_{ab} = T_{(ra)(rb+d)}$$

I don't see how it's possible for the right cosets to be left cosets.

Instead of $T$, I will use $f$ so that it is easier to distinguish between the function and the subgroup.

Ques: When are two right cosets the same?

Suppose $Hf_{ab}=Hf_{cd}$. Then $f_{ab} circ f_{cd}^{-1} =f_{left(frac{a}{c}right), left(b-frac{ad}{c}right)} in H$. This means $frac{a}{c} in Bbb{Q}$. For example, this means for a given $a neq 0$ and regardless of the values taken by $b$ and $c$, both the functions $f_{ab}$ and $f_{ac}$ will be in the same right coset ($because , frac{a}{a}=1 in Bbb{Q}$).

Thus for $f_{ab} in G$, the corresponding right coset will look like
$$Hf_{ab}={f_{cd} in G , | , c=ra text{ for some } r in Bbb{Q}-{0}}.$$

Similarly we deal with the left cosets. To get
$$f_{ab}H={f_{cd} in G , | , a=rc text{ for some } r in Bbb{Q}-{0}}.$$
$color{blue}{text{NOTE:}}$ the only difference is this time the condition is $frac{c}{a} in Bbb{Q}$.

However $frac{a}{c} in Bbb{Q}$ and $frac{c}{a} in Bbb{Q}$ are equivalent conditions (as long as $ac neq 0$). Thus the two cosets are same.

The group of affine transformations, or functions $mathbb{R} to mathbb{R}$ of the form $ x mapsto ax + b $ are isomorphic to a group of $2 times 2$ real matrices of the form $ left[begin{array}{cc} a & b \ 0 & 1 \ end{array} right]$ where $a ne 0$ .

For ease of notation, let's call the group of matrices $G$ for the purposes of this answer. It's isomorphic to the $G$ of the question.

The left and right cosets of a subgroup $H subseteq G$ are equal if and only if the subgroup $H$ is normal in $G$ .

$H$ is normal in $G$ if and only if $H$ is self-conjugate.

To show that $H$ is self-conjugate, let's consider an arbitrary member of $H$, as you have done.

$$ left[begin{array}{cc} r & d \ 0 & 1 end{array}right] tag{1} in H ;;;; text{ where $r in mathbb{Q}$ and $r ne 0$} $$

and let's conjugate it by arbitrary member of $G$, $ left[begin{array}{cc} a & b \ 0 & 1 end{array}right] $, with only the restriction $a ne 0$ imposed.

$$ left[begin{array}{cc} a & b \ 0 & 1 \ end{array}right] left[begin{array}{cc} r & d \ 0 & 1 end{array}right] left(left[begin{array}{cc} a & b \ 0 & 1 end{array}right]^{-1}right) tag{2a} $$

$$ left[begin{array}{cc} r & -rb + b + ad \ 0 & 1 end{array}right] tag{2b} $$

If (1) is in $H$, then (2b) is in $H$ because $r$ is rational and not equal to zero and $-rb + b + ad$ is real.

If (2b) is in $H$, then (1) is in $H$.

By assumption
$$-rb + b + ad in mathbb{R} tag{3a} $$
$a ne 0$, thus
$$frac{-rb}{a} + frac{b}{a} + d in mathbb{R} tag{3b}$$
$frac{rb}{a}$ is real and $frac{-b}{a}$ is real.
$$ d in mathbb{R} tag{3c} $$

therefore $H$ is self-conjugate, therefore $H$ is normal in $G$, therefore the collection of its left cosets is equal to the collection of its right cosets.