Rotation Group and Lorentz Group

One can embed the $3\times3$ rotation matrices

$$R~\in~ SO(3)~:=~\{R\in{\rm Mat}_{3\times 3}(\mathbb{R}) \mid R^tR~=~{\bf 1}_{3\times 3}~\wedge~ \det(R)=1 \}$$

into the $4\times4$ Lorentz matrices

$$\Lambda~\in~ O(1,3)~:=~\{\Lambda\in{\rm Mat}_{4\times 4}(\mathbb{R}) \mid\Lambda^t\eta \Lambda~=~\eta \}$$

as

$$SO(3)~\ni~R~\stackrel{\Phi}{\mapsto}~ \Lambda ~=~ \left[\begin{array}{cc}1 & 0 \cr 0 &R \end{array} \right]~\in~ O(1,3).$$

It is not hard to see that this embedding $\Phi:SO(3)\to O(1,3)$ is an injective group homomorphism

$$\Phi(R_1 R_2)=\Phi(R_1)\Phi(R_2), \qquad R_1,R_2~\in~SO(3).$$

The pertinent group operations are for both groups just matrix multiplication.

Yes, this is a result rigorously stated as: There's a proper subgroup of $O(1,3)$ isomorphic to $SO(3)$. It's made up of the set of Lorentz transformations of the form:

$$\left(\begin{array}{cc} 1 & 0\\ 0 & R(3) \end{array}\right)$$

where $R(3)\in SO(3)$,

together with the internal operation of matrix multiplication. Why is this relevant? Well, to assess that some relevant topological properties of the Lorentz group are inherited from the 3-dimensional proper rotation group.