 Solution

Write $B(x)$ for the displayed matrix. The hyperbolic addition identities giveThus matrix multiplication is closed in $G$. It is associative because ordinary matrix multiplication is associative. The identity is $B(0)=I$, and $B(x)B(-x)=I$ gives the inverse $B(x{)}^{-1}=B(-x)$. Hence $\boxed{G\text{ is a group}}$.

Moreover, $x\mapsto B(x)$ is an injective homomorphism from the additive real numbers, since $\sinh x$ determines $x$. Therefore $\boxed{G\cong (\mathbb{R},+)}$; in particular it is abelian. These matrices are the connected boost subgroup relevant to the component quotient of the real Lorentz group in two dimensions.