Proof that Kronecker's Delta is invariant under Lorentz transformation

Question

There's an exercise in my book that says

"Prove that the Kronecker $\delta$ is invariant under Lorentz transformations".

The solution says that from the property $$\Lambda^Tg\Lambda=g$$ of Lorentz transformations follows that $$\delta^{\mu}_{\nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\beta}_{\nu}\delta^{\alpha}_{\beta}$$but I don't understand why.

From the first relation if we multiply for $g$ at right, since the metric tensor $g$ is the inverse of itself, we obtain $$\Lambda^Tg\Lambda g=I$$ which means that $$(\Lambda^T)^{\mu}_{\alpha}g_{\alpha\beta}\Lambda^{\beta}_{\gamma}g_{\gamma\nu}=\delta ^{\mu}_{\nu}$$ or $$\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\gamma}g_{\alpha\beta}g_{\gamma\nu}=\delta ^{\mu}_{\nu}$$ but I dont's see why $$\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\gamma}g_{\alpha\beta}g_{\gamma\nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\beta}_{\nu}\delta^{\alpha}_{\beta}$$

Answer 1

Just use the fact that $g^\alpha_\beta = \delta^\alpha_\beta$. So in components: \begin{equation} g_{\mu \nu} =\Lambda^\alpha_\mu \Lambda^\beta_\nu g_{\alpha \beta} \Longleftrightarrow \delta^\mu_\nu = \Lambda_\alpha^\mu \Lambda_\nu^\beta \delta_\beta^\alpha \end{equation}