The following theorem was proved in [Helmut Wielandt. Eine Verallgemeinerung der invarianten Untergruppen. Mathematische Zeitschrift 45 (1939): 209-244.] a long time ago:
Theorem: (Wielandt) There exists a function $f: \mathbb N \to \mathbb N$, such that for every finite group $G$ and a sub-normal subgroup $H<G$ such that $C_G(H)=\{1\}$, we have $|G| \leq f(|H|).$
Here, $C_G(H)$ denotes the centralizer of $H$ in $G$. Recall, a subgroup $H<G$ is called subnormal if there is a chain of subgroups $$H=G_0 < G_1 < \dots <G_n=G,$$ such that $G_i$ is normal in $G_{i+1}$.
The theorem implies among other things that the automorphism tower $${\rm Aut}^0(G)=G, \quad {\rm Aut}^{n-1}(G) \to {\rm Aut}^n(G):= {\rm Aut}({\rm Aut}^{n-1}(G))$$ of a centerless finite group terminates after finitely many steps, where bounds on number of steps can be obtained from bounds on the mysterious function $f$. Since Wielandt's theorem is more general, it is unclear (to me) if there could be a different method to obtain better bounds on the number of steps needed. However, let us focus on Wielandt's function $f$ for now:
Question: What are the best known lower and upper bounds for Wielandt's function $f$?
One upper bound, which is of the form $$f(n) = \exp(O(n \log(n))),$$ is coming from the following result. It was proved in [Martin Pettet, A note on the automorphism tower theorem for finite groups, Proc. Amer. Math. Soc. 89 (1983), 182-183], that in the situation above $$|G| \leq |Z(H_{\infty})| \cdot |{\rm Aut}(H_{\infty})|!,$$ where $H_{\infty}$ denotes the nilpotent residual, i.e., the smallest normal subgroup, such that $H/H_{\infty}$ is nilpotent, and $Z(H_{\infty})$ denotes the center of $H_{\infty}$.
A lower bound for $f$ could come from something like $H^n < H^n \rtimes {\rm Sym}(n)$, for some simple group $H$ like ${\rm Alt(5)}$ which yields $$f(60^n) \geq 60^n \cdot n!$$ or in other words $$f(n) = \exp(\Omega(\log(n)\log\log(n)) = n^{\Omega(\log\log(n))}.$$
There is a substantial gap between these lower and upper bounds. Can anyone close the gap?
