Index notation problem in getting $(\Lambda^{-1})^{\mu}_{\nu}=g^{\mu\alpha}g_{\nu\beta}\Lambda^{\beta}_{\alpha}$

Question

Let consider:

$g_{\mu\nu}=diag(1,-1,-1,-1)$ and $\Lambda$ so that $\Lambda^{T}g$$\Lambda$=g

I want to prove that $g=g\Lambda\Lambda \Rightarrow \Lambda^{-1}=g^{-1}g\Lambda$

The proof is silly for me with matrix notation:

$g=g\Lambda\Lambda$ $\longrightarrow$ $g\Lambda^{-1}=g\Lambda$ $\longrightarrow$ $\Lambda^{-1}=g^{-1}g\Lambda$

I don't manage with it in index notation:

$g_{\mu\nu}=g_{\alpha\beta}\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\nu}$ $\,\, \Longrightarrow \,\, (\Lambda^{-1})^{\mu}_{\nu}=g^{\mu\alpha}g_{\nu\beta}\Lambda^{\beta}_{\alpha}$

I tried to do the same passages:

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

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

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

Now I' m stuck.I know that my problem is that I am not able to do calculations in index notation. Someone can share a step-by-step solution so I can learn the rules I'm doing wrong.

Answer 1

Passing e.g. from $g=\Lambda^Tg\Lambda$ to $g=g\Lambda^2$ is nonsense; you need only think in terms of matrices, not tensors, to see this. Given the way matrix multiplication places matching indices next to each other, the correct way to translate the desired result into matrix notation is $\Lambda^{-1}=g^{-1}\Lambda^T g$. This is a trivial consequence of $\Lambda^T g\Lambda=g$ (provided $\det\Lambda\ne 0$; luckily, we can prove $\det\Lambda=\pm 1$ using $\det g\ne 0$.)