Solution
= Solution
With $\eta=\operatorname{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional <Lorentz group> is
$$
\Lambda^T\eta\Lambda=\eta.
$$
Taking <determinant>[determinants] gives $(\det\Lambda)^2=1$, hence $\det\Lambda=\pm1$. The norm of the zeroth column gives
$$
(\Lambda^0{}_0)^2-(\Lambda^1{}_0)^2-(\Lambda^2{}_0)^2=1,
$$
so $|\Lambda^0{}_0|\geq1$. The <Proper orthochronous Lorentz group> is the component with $\det\Lambda=1$ and $\Lambda^0{}_0\geq1$.