I really hope someone can help with that.. $ c>0 \in \mathbb{R}. \mathbb{R^2}$ is a vector space with a symmetric form s with $ s((x,t),(x',t')) := xx'-c^2tt'$.
to Show is that all elements positive determinants in $ G(\mathbb{R^2},s)$ are fulfilled by matrices in the shape of $ +- L_v$ with
$L_v:=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}* \begin{pmatrix} 1 & -v \\ -\frac{1}{c^2}*v & 1\\ \end{pmatrix} $
$v \in \mathbb{R}$ and $|v| < c $
looking Forward to your help! :)
Information about the generators of the Lorentz group is found elsewhere, I'm just going to give you a brief sketch of the steps you need to follow for the $1+1$ case. A Lorentz transformation $\Lambda(v)$ is defined as the element of the general group such that
$$ \Lambda(v)^T \eta \Lambda(v) = \eta ~~~\mbox{with}~~~ \eta = \left(\begin{array}{cc}1 & 0 \\ 0 & -1 \end{array}\right) \tag{1} $$
If you take the determinant of this equation you find that
$$ |\det{\Lambda(v)}|^2 = 1 ~~~\Rightarrow \det \Lambda(v) = \pm 1 $$
I will just consider the case $\det \Lambda(v) = +1$. Imagine that $\epsilon$ is a very small number, so that you can express $\Lambda$ as
$$ \Lambda(\epsilon) = 1 + \epsilon S + \mathcal{O}(\epsilon^2) \tag{2} $$
This is basically a Taylor expansion on powers of $\epsilon$. If you replace Eq. (2) in Eq. (1) and keep all terms up to linear order you get the constraint
$$ S^T\eta + \eta S = 0 \tag{3} $$
To find all matrices that satisfy this, imagine the most general matrix you can write
$$ S = \left(\begin{array}{cc}a & b \\ c & d \end{array}\right) $$
Replacing this in Eq. (4) you can conclude that $a = d = 0$ and $b = c$, which then leads to
$$ S = \left(\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array}\right) \tag{4} $$
without loss of generality I assume that $b=1$, since the constants can be absorbed by $\epsilon$.
With this at hand, you can calculate any matrix $\Lambda(u)$ for any value of $u$, just split it into $N$ pieces and use Eq. (2)
$$ \Lambda(u) = \underbrace{\left(1 + \frac{u}{N}S\right)\cdots \left(1 + \frac{u}{N}S\right)}_{N~{\rm times}} = \left(1 + \frac{u}{N}S\right)^N \stackrel{N\to \infty}{=} e^{uS} \tag{5} $$
The exponential of this matrix is easy to obtain,
$$ \Lambda(u) = \left(\begin{array}{cc}\cosh u & \sinh u \\ \cosh u & \sinh u \end{array}\right) \tag{6} $$
If you call $\beta = \tanh u = v/c$, and $\gamma = \cosh u$ you get to the expression you are looking for.