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I am studying the case of infinitesimal proper Lorentz transformations, and i stumble at this notation result : $\Delta\omega^\alpha_\gamma g_{\alpha\beta} g^\beta_\delta = \Delta \omega_{\delta\gamma}$?

I know that the following is true : $x^\mu = g^{\mu\sigma} x_\sigma$. So what happens here is that the metric tensor raises the index and then the raised index changes to the Greek letter which is not contracted, μ in this case. How does the same applies in the above equation which i cannot understand ?

Edit : Prahar was correct i had made a mistake. I solved it and this is how :

$g_{\alpha\beta} g^\beta_\delta = g_{\alpha\delta}$

$\Delta\omega^\alpha_\gamma*g_{\alpha\delta}=\Delta\omega_{\delta\gamma}$

I am studying the case of infinitesimal proper Lorentz transformations, and I stumble at this notation result : $\Delta\omega^\alpha_\gamma g_{\alpha\beta} g^\beta_\delta = \Delta \omega_{\delta\gamma}$?

I know that the following is true : $x^\mu = g^{\mu\sigma} x_\sigma$. So what happens here is that the metric tensor raises the index and then the raised index changes to the Greek letter which is not contracted, μ in this case. How does the same applies in the above equation which I cannot understand ?

Edit : Prahar was correct I had made a mistake. I solved it and this is how :

$g_{\alpha\beta} g^\beta_\delta = g_{\alpha\delta}$

$\Delta\omega^\alpha_\gamma\cdot g_{\alpha\delta}=\Delta\omega_{\delta\gamma}$