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 $$\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\gamma}g_{\alpha\beta}g_{\gamma\nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\beta}_{\nu}\delta^{\alpha}_{\beta}$$