# Lie groups with same algebra

The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1.

In general, you need to know the Lie group itself to find the correct subgroup (i.e. you can't just find the subgroup from the algebra alone). This is exactly because of cases like $SU(2)$ and $SO(3)$ that have isomorphic tangent spaces, but which have different global properties.