For the mapping x to ax+b, show that every left coset is a right coset I'm trying to answer the following question from Herstein: Topics in Algebra (1st) section 2.5 Question 7-9.
Given the mapping:
$$T_{ab} : x\rightarrow ax+b $$
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.
 A: 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.
A: 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. 
