= Solution
Write $X(\varphi)=\varphi K$. Since $K^2=I$, the <Exponential map of a Lie group> gives
$$
\Lambda(\varphi)=e^{\varphi K}
=\cosh\varphi,I+\sinh\varphi,K
=\begin{pmatrix}\cosh\varphi&\sinh\varphi\\\sinh\varphi&\cosh\varphi\end{pmatrix}.
$$
The hyperbolic addition formulas give $\Lambda(\varphi)\Lambda(\psi)=\Lambda(\varphi+\psi)$ and $\Lambda(\varphi)^{-1}=\Lambda(-\varphi)$, 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 $\cosh\varphi$ is unbounded, equivalently because the subgroup is homeomorphic to $\mathbb R$.
Solved by gpt-5.6-sol high.
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