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Since $X^3=0$,Every group element has the unique preimage $t=s-qr/2$, so the Exponential map of a Lie group is bijective. It is a diffeomorphism from ${\mathbb{R}}^3$ onto $G$, proving that $G$ is connected and simply connected.

Write $X(\phi )=\phi K$. Since $K^2=I$, the Exponential map of a Lie group givesThe hyperbolic addition formulas give $\Lambda (\phi )\Lambda (\psi )=\Lambda (\phi +\psi )$ and $\Lambda (\phi {)}^{-1}=\Lambda (-\phi )$, so these matrices form a subgroup of the One-dimensional Lorentz group. It is Abelian because addition in $\mathbb{R}$ is commutative. It is noncompact because $\cosh \phi$ is unbounded, equivalently because the subgroup is homeomorphic to $\mathbb{R}$.

For $\phi \ne \pm 1$, the Cayley transform isWriting its diagonal and off-diagonal entries as $a,b$, one has $a^2-b^2=1$, so $C^T\eta C=\eta$ and $\det C=1$. For $|\phi |<1$, put $\phi =\tanh (t/2)$ to obtain $C(\phi K)=\Lambda (t)$, so this interval covers the identity component. For $|\phi |>1$ the image lies in the other component and covers it except for $-I$, approached only as $|\phi |\to \infty$. The Exponential map of a Lie group reaches only the identity component, whereas the Cayley transform also reaches nonidentity-component elements but omits $-I$ and is undefined at $\phi =\pm 1$.