Find the exponential map $\operatorname {exp}_G : \mathfrak g \to G$? Let $
Q=\left(\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right) \text { and } G:=\left\{A \in G L(2, \mathbb{R}) / A^{t} Q A=Q\right\}
$
Calculate explicitly the exponential map $\operatorname {exp_G} : \mathfrak g \to G$ ? Is it onto ?

I found that the Lie algebra $\mathfrak g$ is $$\mathfrak g=\{A \in M_2(\mathbb R) : (QA)^{T}=-QA\}$$
or $G$ is a matrix group, so $\operatorname{exp}_G=\operatorname{exp} :M_2(\mathbb R) \to GL_2(\mathbb R)$ defined by :
$$ exp(A)=\sum_{k=0}^{\infty} \frac{A^k}{k!}$$
that is onto by construction.
Is my answer correct ? Any help is appreciated !
 A: The exponential map $\exp_G: \mathfrak{g}\to G$ isn't onto. For example $Q\in G$ (obvious) but $Q\notin\exp(\mathfrak{g})$ (e.g. because for $M\in\mathfrak{g}$, $\det(\exp(M))=\exp(tr(M))>0$; yet $\det(Q)=-1$).
(And btw, be careful with the definition of $\exp$ : the sum starts at $k=0$, not $k=1$)
A: $G$ consists of 4 disjoint parts:
$$\begin{align}
G_{++} &= \left\{ \begin{pmatrix}
\cosh s & \sinh s \\
\sinh s & \cosh s
\end{pmatrix} : s\in\mathbb{R} \right\} \ni I
\\
G_{+-} &= \left\{ \begin{pmatrix}
\cosh s & \sinh s \\
-\sinh s & -\cosh s
\end{pmatrix} : s\in\mathbb{R} \right\} = \{ Qg : g\in G_{++} \} \ni Q
\\
G_{-+} &= \left\{ \begin{pmatrix}
-\cosh s & -\sinh s \\
\sinh s & \cosh s
\end{pmatrix} : s\in\mathbb{R} \right\} = \{ -Qg : g\in G_{++} \} \ni -Q
\\
G_{--} &= \left\{ \begin{pmatrix}
-\cosh s & -\sinh s \\
-\sinh s & -\cosh s
\end{pmatrix} : s\in\mathbb{R} \right\} = \{ -g : g\in G_{++} \} \ni -I
\end{align}$$
Only $G_{++}$ is generated by the exponential map from the Lie algebra which is 1-dimensional:
$$
\mathfrak{g}=\left\{ \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} : t\in\mathbb{R} \right\}
$$
