# Isometries in Minkowski space

Consider theorem 1.7 from chapter III of 'Elementary differential geometry' by O'Neill. It says that:

Theorem 1.7: If $$\phi$$ is an isometry of $$E^3$$, then there exists a unique translation $$T$$ and a unique orthogonal transformation $$\psi$$ such that $$\phi = T\psi$$.

Now, consider the relativistic interval defined on $$R^n$$ as usual, i.e. $$I(x,y) = \sqrt{\sum (x_i - y_i)^2 - c^2(x_n - y_n)^2}$$. We say that $$\phi$$ is a minkowskian isometry from $$R^n$$ to $$R^n$$ just in case for any x,y $$\in R^n$$, $$I(x,y) = I(\phi(x), \phi(y))$$.

The question is: does the theorem 1.7 transfer immediately to the minkowskian isometry? That is, can we decompose the Minkowskian isometry into a translation $$T'$$ in Minkowski space and an orthogonal transformation $$\psi'$$ (the Lorentz transformation) ?