Solution: FFirst of all note that for any $T_{a,b}, T_{c,d} \in G$, we have \begin{align*} (T_{a,b} \circ T_{c,d})(r) & = T_{a,b}(T_{c,d}(r)) \\ & = T_{a,b}(cr + d) \\ & = a(cr+d) + b = acr + ad + b \\ & = T_{ac, ad+b}(r). \end{align*} Now we need to show that $H$ is a group under the above operation. It is clear that if $a,c \in \mathbb{Q} $, then $ac \in \mathbb{Q} $ and hence $T_{a,b}, T_{c,d} \in H $ implies $T_{a,b} \circ T_{c,d} \in H$.

Associativity also follows as \begin{align*} (T_{a,b} \circ T_{c,d}) \circ T_{m,n} = T_{ac, ad + b} \circ T_{m,n} = T_{amc, acn + ad + b} \\ T_{a,b} \circ (T_{c,d} \circ T_{m,n}) = T_{a,b} \circ T_{cm, cn + d} = T_{amc, acn+ad+b}. \end{align*}

We claim that $T_{1,0}$ is the identity. Note that for any $T_{a,b}\in H$ \begin{align*} T_{1,0}\circ T_{a,b} = T_{a,b} \quad \text{ and } T_{a,b} \circ T_{1,0} = T_{a, b}. \end{align*} Finally, for the inverse of $T_{a,b}$ let $T_{x,y}$ be the inverse, then \begin{align*} T_{a,b} \circ T_{x,y} = T_{1,0} & \implies T_{ax, ay + b} = T_{1,0} \\ & \implies ax = 1 \text{ and } ay + b = 0 \\ & \implies x = \frac{1}{a} \text{ and } y = -\frac{b}{a}. \end{align*} Since $a \in \mathbb{Q}-\{ 0 \} $, so is $a^{-1} $. Thus, the inverse of $T_{a,b}$ will be $T_{a^{-1} , -a^{-1}b}$. This is a simple verification as \begin{align*} T_{a,b} \circ T_{\frac{1}{a},-\frac{b}{a}} = T_{1,0}. \end{align*}

Now we claim that $H$ is non-abelian. Note that \begin{align*} T_{c,d} \circ T_{a,b} = T_{ca, cb + d}. \end{align*} Thus, $H$ will be abelian if $cb + d = ad + b$, which is not true in general. For example, $a = b = 1$ and $c = 2, d = 3$. Thus, $H$ is nonabelian.