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.
Best Answer
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.