 Solution

The identity matrix belongs to $O(1,3)$. If $A,B\in O(1,3)$, thenso the set is closed under multiplication. Taking determinants in $A^{T}MA=M$ gives $(\det A{)}^2=1$, so $A$ is invertible. Rearranging the defining identity givesIts inverse also preserves $M$, either directly or becauseThus the Lorentz group $O(1,3)$ is a group under matrix multiplication.