Solution
= Solution
The identity matrix belongs to $O(1,3)$. If $A,B\in O(1,3)$, then
$$
(AB)^TM(AB)=B^T(A^TMA)B=B^TMB=M,
$$
so the set is closed under multiplication. Taking determinants in $A^TMA=M$ gives $(\det A)^2=1$, so $A$ is invertible. Rearranging the defining identity gives
$$
A^{-1}=M^{-1}A^TM=MA^TM.
$$
Its inverse also preserves $M$, either directly or because
$$
(A^{-1})^TM A^{-1}=M.
$$
\b[Thus the <Lorentz group> $O(1,3)$ is a group under matrix multiplication.]