Recently I explored that SGH catches FGH not even at LVO, but even at Feferman-Schutte ordinal. I noted that each increment to the FGH ordinal leads to the same growth rate in the SGH ordinal, but when that increment is placed inside theta function. Only thing what we have to prove is that if
\(g_{\theta(\alpha)}(n) \approx f_{\alpha}(n)\)
Then:
\(g_{\theta(\alpha,m)}(n) \approx f_{\alpha}(n(m+1))\)
From that we could find that \(g_{\theta(\alpha+1)}(n) \approx f_{\alpha+1}(n)\), as \(g_{\theta(\alpha+1)}(n)\) collapses to \(g_{\theta(\alpha,\theta(\alpha,\cdots,\theta(\alpha)\cdots))}(n)\) (n \(\alpha\)'s) and \(f_{\alpha+1}(n)\) to \(f_\alpha(f_\alpha(f_\alpha(\cdots(f_\alpha(n))\cdots))\) (n \(\alpha\)'s).
So, roughly speaking, we just need to remove one \(\theta()\) from the resulting SGH ordinal in order to get to the corresponding FGH ordinal. Then it is obvious why FGH catches SGH namely on \(\Gamma_0 = \theta(\theta(\theta(\cdots)))\)
Update[]
I must admit that my research was wrong starting from approximation \(g_{\theta(\omega,1)}(n)\) as \(f_\omega(2n)\). In fact in turned out that \(g_{\theta(\omega,1)}(n)\) is roughly \(f_n(f_n(n))\) and \(g_{\theta(\omega,m)} \approx f_n(f_n(n)*m)\). By that it is clear that \(g_{\theta(\omega,\theta(\omega,1))} \approx f_n(f_n(f_n(n)))\).
Then \(g_{\theta(\omega+1)}(n) \approx f_{n+1}(n)\). We see that operations with \(\omega\)'s inside theta function in SGH work like operations with n's in FGH. It follows that:
\(g_{\theta(\omega*2)}(n) \approx f_{2n}(n)\)
\(g_{\theta(\omega^2)}(n) \approx f_{n^2}(n)\)
\(g_{\theta(\omega^\omega)}(n) \approx f_{n^n}(n)\)
\(g_{\theta(\theta(1))}(n) \approx f_{f_3(n)}(n)\)
\(g_{\theta(\theta(2))}(n) \approx f_{f_4(n)}(n)\)
\(g_{\theta(\theta(\omega))}(n) \approx f_{f_n(n)}(n)\)
We see that \(g_{\Gamma_0}(n) = g_{\theta(\theta(\theta(\cdots)))}(n)\) will correspond to n nested number subscripts, which is equivalent to \(f_{\omega+1}(n)\).