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It is often stated that rotations in the 3 spatial dimensions are examples of Lorentz transformations.
But Lorentz transformations form a group named the Lorentz Group, $O(1,3)$ which is a group a $4 \times 4$ matrices, $\Lambda$ having the following property:
$$ \Lambda^T g \Lambda = g$$
where $g$ is the metric tensor.
Now rotations matrices for the 3 spatial dimensions are $3 \times 3$ matrices and form $SO(3)$. How can they be in the $O(1,3)$ ?
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.
