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There is a correspondence between the Lorentz group and the group $SL(2,\mathbb{C})$. To each Lorentz transformation $\Lambda$ we can associate two matrices $\pm A(\Lambda) \in SL(2,\mathbb{C})$ such that $$A \sigma_{\nu} A^{\dagger} = \Lambda^{\mu}_{\ \ \nu} \ \sigma_{\nu},$$ where $\sigma_{\nu}$ ($\nu = 1,2,3$) are the Pauli matrices. I was trying to prove that the matrices $\{A(\Lambda)\}$ forms a representation of the Lorentz group, and I would like if my proof is correct.

Since the product of two Lorentz transformations $\Lambda_{1}$ and $\Lambda_{2}$ is a Lorentz transformation ($\Lambda$), using the relation above, we have $$A(\Lambda) \sigma_{\nu} A^{\dagger}(\Lambda) = \Lambda^{\mu}_{\ \ \nu} \ \sigma_{\nu} = \Lambda^{\mu}_{1 \ \alpha} \Lambda^{\alpha}_{2 \ \nu} \ \sigma_{\nu} \\ = (\Lambda^{\mu}_{1 \ \alpha}\sigma_{\nu}) \Lambda^{\alpha}_{2 \ \nu} $$ (using that $\Lambda_{1}$ also satisfy the first equation) $$ = A(\Lambda_{1}) \sigma_{\alpha} A^{\dagger}(\Lambda_{1})\Lambda^{\alpha}_{2 \ \nu} \\ = A(\Lambda_{1}) (\Lambda^{\alpha}_{2 \ \nu} \sigma_{\alpha}) A^{\dagger}(\Lambda_{1}) \\ = A(\Lambda_{1}) A(\Lambda_{2}) \sigma_{\nu} A^{\dagger}(\Lambda_{2}) A^{\dagger}(\Lambda_{1}).$$ So, in the end I get $$A(\Lambda) \sigma_{\nu} A^{\dagger}(\Lambda) = A(\Lambda_{1}) A(\Lambda_{2}) \sigma_{\nu} (A(\Lambda_{1}) A(\Lambda_{2}))^{\dagger}.$$ Can I conclude from this that $A(\Lambda) = A(\Lambda_{1})A(\Lambda_{2})$?

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