Chain of automorphism groups

Question

Let $\mathrm{Aut}(\Gamma)$ be the automorphism group of a graph $\Gamma$. Also suppose that $\mathrm{Cay}(G,S)$ is the Cayley graph of a group $G$ with respect to the generating set $S$. Consider the following chain:

$\Gamma \to \mathrm{Aut}(\Gamma) \to \mathrm{Cay}(\mathrm{Aut}(\Gamma),S_1) \to \mathrm{Aut}(\mathrm{Cay}(\mathrm{Aut}(\Gamma),S_1)) \to \mathrm{Cay}(\mathrm{Aut}(\mathrm{Cay}(\mathrm{Aut}(\Gamma),S_1)),S_2) \to \ldots$ ,

where arrow $A\to B$ just means making $B$ from $A$. For example $\Gamma \to \mathrm{Aut}(\Gamma)$ means that the automorphism group $\mathrm{Aut}(\Gamma)$ is constructed based on $\Gamma$. Is there any graph $\Gamma$ for which the above chains stops, i.e. after $i$ steps

$\Gamma \cong \mathrm{Cay}(\mathrm{Aut}(\mathrm{Cay}(\cdots(\mathrm{Aut}(\Gamma))),S_i)$,

as two graphs (the notation $\cong$ means graph isomorphism)?

What about the case that starts with a group $G$, i.e. is there any group $G$ such that

$G \cong \mathrm{Aut}(\mathrm{Cay}(\mathrm{Aut}(\cdots(\mathrm{Cay}(G,S_1)),S_i))$,

as two groups (the notation $\cong$ means group isomorphism and $\mathrm{Aut}(G)$ is the automorphism group of the group $G$ here)?

Answer 1

The question is not well-defined, you need to pick a generating set for each group. (Sometimes, there are only infinite generating sets.) I'll interpret the second question as

"For which groups $G_0$ does there exist a sequence $(G_i,S_i)_{0\le i\le n}$ of groups with (possibly infinite) generating sets such that $ G_{i+1} = \mathrm{Aut}(\mathrm{Cay}(G_i,S_i))$ and $G_n\simeq G_0$?"

If we suppose $G$ is finitely generated, then we can answer Question 2 pretty easily using results of de la Salle and Leemann. The first observation is that $$G_i\le \mathrm{Aut}(\mathrm{Cay}(G_i,S_i)) = G_{i+1}$$ for any generating set.

A) If $G_0$ does not belong to the list in Corollary 1.2 of https://arxiv.org/pdf/2010.06020, then there exists a generating set such that $G_0=\mathrm{Aut}(\mathrm{Cay}(G_0,S_0))$ so the chain loops pretty quickly.

B) If $G_0$ belongs to the list, then $G_0<G_1$. There are multiple cases

• If $G_0$ is finite, $G_0$ cannot be a proper subgroupf of itself, we are done.
• If $G_0$ is abelian and infinite, then $G_1$ is non-abelian, and this remains true for all higher $G_i$. (The automorphism $x\mapsto x^{-1}$ doesn't not commute with the translation by an element of infinite order.)
• If $G_0$ is generalized dicyclic and infinite, meaning there exists an index-two abelian group $A$ and an element $x\notin A$ such that $x^4=1$ and $xax^{-1}=a^{-1}$ for all $a\in A$, then the (graph) automorphism $$ g\mapsto \begin{cases} g & \text{ if }g\in A \\ g^{-1} & \text{ if }g\notin A \end{cases} $$ does not commute with translation by $A$, so $[G_1:H]\ge 4$ for all abelian subgroups $H$, and this remains true for all higher $G_i$.

So in all those cases, we cannot come back to $G_0$. (However, we can ensure the chain stabilizes. If we are careful, we can take $[G_{i+1}:G_i]\le 2$ by https://arxiv.org/abs/2105.02326, so all $G_i$ are finitely generated, and eventually we can apply part A).)