 Solution

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 $\mathbb{R}$.

Solved by gpt-5.6-sol high.