Deriving Lie Algebra for a Lie Group I'm reading Helgason's DG, Lie Groups, and Symmetric Spaces and at one point he briefly mentions the Lobatchevski half-plane on page 136:

The group $G$ of the mappings $T_{a,b}: x \to ax+b$ with $x \in \mathbb{R}$ and $a>0$ has a Lie algebra $\mathfrak{g} = \mathbb{R}e_1 + \mathbb{R}e_2$ where $[e_1,e_2] = e_2$.

I'm trying to verify the claim that the bracket is what it is by taking $[\cdot, \cdot] \equiv d\left (Ad\right )_e:\mathfrak{g\times g\to g}$ but I'm having trouble verifying this. We can see that 
$$
Ad_{a,b} \circ T_{c,d} \;\; =\;\; T_{c,\; ad-bc+b}
$$
If $\gamma, \eta$ are curves such that $\gamma(0) = \eta(0) = e$ and $\dot{\gamma}(0) =X, \; \dot{\eta}(0) = Y$ then we obtain
\begin{eqnarray*}
[X,Y] & = & ad_X(Y) \\
& = & \left. \frac{d}{dt} \right |_{t=0} Ad_{\gamma(t)} \circ T_{\eta(t)} \\
& = & \left . \frac{d}{dt} \right |_{t=0} \left (c(t), \; a(t)d(t) - b(t)c(t) + b(t) \right ) \\
& = & \left (\dot{c}(0), \; \dot{a}(0)d(0) + a(0)\dot{d}(0) - \dot{b}(0)c(0) - b(0)\dot{c}(0) + \dot{b}(0)\right ) \\
& = & \left (Y_1, Y_2\right ) \;\; =\;\; Y
\end{eqnarray*}
where I make the short hand $\gamma(t) = (a(t), b(t)), \; \eta(t) = (c(t), d(t))$ and $X = (X_1, X_2), \; Y = (Y_1,Y_2)$.
On first inspection it looks as if this computation satisfies the bracket, but I'm not satisfied with it since this is done independent of a basis for $\mathfrak{g}$.  Can anyone point out what I'm missing or what's wrong with this computation/approach?
 A: There are several ways to do this. The first one is more to my liking, but the second one closest to your computations.
1) One way to compute the Lie Bracket, assuming we know what the Lie bracket on $\mathfrak{gl}(n)$ is,  is to work through a linear representation. Let $\Psi : G \to GL(2)$ be the following map:
$$\Psi(T_{a,b})=\left( \begin{array}{cc}  a & b \\ 0 & 1 \end{array} \right).$$
This map is an injective morphism, that allows to identify $G$ with its image.
Thus, $\mathfrak{g}$ is identifed with the subalgebra
$$\mathfrak{g}\simeq \left\{ \left(\begin{array}{cc}  * & * \\ 0 & 0 \end{array} \right) \right\}, $$
and the Lie bracket in this representation is the usual Lie bracket on $\mathfrak{gl}(n)$ - this comes from the fact this is true for the group $GL(n)$, which contains our group. 
Put 
$$e_1=\left(\begin{array}{cc}  1 & 0 \\ 0 & 0 \end{array} \right), \; e_2=\left(\begin{array}{cc}  0 & 1 \\ 0 & 0 \end{array} \right),$$
this is clearly a basis and then
$$[e_1,e_2]=e_1e_2-e_2e_1=e_2,$$
by direct matrix product computations.
2) As noted by Dap in comments, you should be careful with definitions; $Ad$ is a map from $G$ to $Aut(\mathfrak{g})$, $Ad_g$ is the derivative of $h\mapsto ghg^{-1}$ at identity, and $ad: \mathfrak{g}\to Der(\mathfrak{g})$ can be identified with the derivative of the latter.
Thus, to correct what you wrote,
$$c_{a,b}(T_{c,d})=T_{c,ad-bc+b},$$
where $c_{a,b}$ is $g\mapsto T_{a,b} \circ g \circ T_{1/a,-b/a}$ is the conjugation by $T_{a,b}$.
Let 
$$e_1=\frac{\partial T_{1+t,0}}{\partial t} (0), e_2=\frac{\partial T_{1,t}}{\partial t} (0).$$
 In particular,
$$c_{a,b}(T_{1+t\alpha,t\beta})=T_{1+t\alpha,t(a\beta-b\alpha)},$$
so when $t\to 0$, we have computed the derivative
$$Ad_{a,b}(\alpha e_1+\beta e_2)=\alpha e_1+ (a\beta-b\alpha)e_2,$$
that is, the matrix of $Ad_{a,b}$ in the basis $(e_1,e_2)$ is
$$Ad_{a,b}=\left( \begin{array}{cc} 1& 0 \\ -b & a \end{array}\right)$$
so the endomorphism $ad_{e_1}$ has matrix in the basis $(e_1,e_2)$
$$\frac{\partial }{\partial t} \left( \begin{array}{cc} 1 & 0 \\ 0 & t \end{array}\right)=\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right),$$
so $[e_1,e_2]=ad_{e_1}(e_2)=e_2$, as expected.
