We embed the rotation group, $SO(3)$ into the Lorentz group, $O(1,3)$ : $SO(3) \hookrightarrow O(1,3)$ and then determine the six generators of Lorentz group: $J_x, J_y, J_z, K_x, K_y, K_z$ from the rotation and boost matrices.
From the number of the generators we realize that $O(1,3)$ is a six parameter matrix Lie group.
But are there any other way to know the number of parameters of the Lorentz group in the first place?
Answer
From special relativity we know that a Lorentz transformation: \begin{equation} x'^\mu = \Lambda^\mu {}_\nu x^\nu \end{equation} preserves the distance: \begin{equation} g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu' \end{equation} The above two equations imply: \begin{equation} g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \end{equation} Now, let us consider an infinitesimal transformation: \begin{equation} \Lambda_\nu {}^\mu = \delta_\nu{}^\mu + \omega_\nu{}^\mu + O(\omega^2) \end{equation} such that we can write: \begin{equation} \begin{aligned} g^{\mu \nu} & = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \\& = g^{\rho \sigma} \left( \delta_\rho{}^\mu + \omega_\rho{}^\mu + \cdots \right)\left( \delta_\sigma{}^\nu + \omega_\sigma{}^\nu + \cdots \right) \\& = g^{\mu \nu} + g^{\mu \sigma} \omega_\sigma{}^\nu + g^{\rho \nu} \omega_\rho{}^\mu + O(\omega^2) \\& = g^{\mu \nu} + \omega^{\mu\nu} + \omega^{\nu \mu} + O(\omega^2) \end{aligned} \end{equation} and so: \begin{equation} \omega^{\mu\nu} = - \omega^{\nu \mu} \end{equation} Thus, the matrix $\omega$ is a $4 \times 4$ antisymmetric matrix, which corresponds to $6$ independent parameters (i.e. the $3$ parameters corresponding to boosts and the $3$ parameters corresponding to rotations).
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