The transfinite tower of iterative automorphisms of a group $G$ is simply definied to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct limit is taken at the limit stages:

$G\rightarrow Aut(G)\rightarrow Aut(Aut(G))\rightarrow\cdots\rightarrow G_{\alpha}\rightarrow G_{\alpha+1}\rightarrow\cdots$

The tower terminates when a fixed point is reached, namely one of the groups in the chain is isomorphic to its automorphism group by the natural map. Simon Thomas has proved that the automorphism tower of every centerless group eventually terminates. Later, Hamkins completed Thomas' result by showing that the automorphism tower terminates for every group:

Hamkins' theorem gives a sense to the natural definition of the notion of terminating number of a group, $\tau(G)$, that is the least ordinal where the automorphism tower of $G$ terminates.

My first question is about the minimum power of $ZFC$ that is needed to carry out Thomas-Hamkins' proof:

Question 1. How much $ZFC$ is needed to prove that the automorphism tower terminates for every group, $G$, and so $\tau(G)$ is well-defined? Particularly, is $AC$ used anywhere in Hamkins or Thomas' results (which Hamkins' proof is partially based on)? If so, is this use of $AC$ essential? If yes, are the following two statements equivalent?

  • The automorphism tower terminates for every group.

  • The Axiom of Choice.

My next question is about the relation between the terminating number of the direct product of two groups and the terminating number of each component:

Question 2. What is the relation between $\tau (G\times H)$ and $\tau (G)$, $\tau(H)$? Is there an upper bound for $\tau (G\times H)$ expressible in terms of $\tau (G)$, $\tau(H)$? For instance, is it true to say $\tau (G\times H)\leq Max (\tau (G), \tau(H))$ or $\tau (G)+\tau(H)$ or $\tau (G).\tau(H)$ ...?

The "Max" bound in the above question is inspired by the fact that for finite groups, $G, H$, whose orders are relatively prime, we have $Aut(G\times H)\cong Aut(G)\times Aut(H)$. If one somehow manages to keep this pattern through the entire chain then the automorphism tower of $G\times H$ terminates after $Max (\tau (G), \tau(H))$ steps.

In particular, computing $\tau(G^n)$ (and comparing it with $\tau(G)$) could be of interest as well. For instance, in the special case that $G$ is a cyclic group of order $p$, one has $Aut(G^n)\cong GL_{n}(\mathbb{F}_p)$ and so $\tau (G^{n})=\tau (GL_{n}(\mathbb{F}_p))+1$.

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    $\begingroup$ My proof that every group leads eventually to a centerless group uses the replacement axiom, since I use that every function from the ordinals to the ordinals has a closure point. I haven't ever thought much about whether the arguments use AC. Hmmm. $\endgroup$ Jul 17, 2018 at 13:33
  • $\begingroup$ @Joel: Every function from $V$ to $V$ has a closure point, assuming Replacement. $\endgroup$
    – Asaf Karagila
    Jul 17, 2018 at 13:41
  • $\begingroup$ Yes, of course, and both forms of closure are equivalent to replacement/collection over separation etc. My proof, however, used explicitly a certain function on the ordinals, the function mapping $\alpha$ to $\beta$, when all the group elements at stage $\alpha$ that will eventually die, die before $\beta$. Meanwhile, it seems that one will want at least power set to define the tower, since otherwise you can't seem to know that the automorphism group even exists as a set. $\endgroup$ Jul 17, 2018 at 13:54
  • $\begingroup$ Any thoughts on the second question? $\endgroup$ Jul 17, 2018 at 14:26
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    $\begingroup$ This is in part what Gunter Fuchs and I had done. In my argument with Simon, we had to force to create the group, but with Gunter, we proved that there is such a malleable group assuming only $\Diamond$. We never got it purely in ZFC. $\endgroup$ Jul 17, 2018 at 16:34

1 Answer 1


Here is a relevant paper,

Kaplan, Itay; Shelah, Saharon, The automorphism tower of a centerless group without choice, Arch. Math. Logic 48, No. 8, 799-815 (2009). ZBL1192.03026.

At least for a centerless group, it seems that every automorphism tower stabilizes below the Lindenbaum number of the power set of finite sequences of the group.

While I don't quite know how to move from a group to a centerless group without choice, let me point out that $G\mapsto\operatorname{Aut}(G)$ is a function on the universe, and by replacement there is some $\alpha$ such that $V_\alpha$ is a closure point of that function. I suspect that an argument would work in a choiceless setting from this point.

  • $\begingroup$ I think my proof does not use AC at all, and so from any group $G$ you get eventually (by the replacement axiom) a centerless group. So this seems to complete the proof in ZF. $\endgroup$ Jul 17, 2018 at 13:58
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    $\begingroup$ @JoelDavidHamkins: Joel, doesn't your proof reduce the general case to the centerless case? If so, Simon's proof, as far as I know, uses AC. $\endgroup$
    – Burak
    Jul 17, 2018 at 14:08
  • $\begingroup$ @Burak: Have you read my answer and at least the abstract of the linked paper? $\endgroup$
    – Asaf Karagila
    Jul 17, 2018 at 14:11
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    $\begingroup$ @AsafKaragila: Yes, the paper proves Simon's result without AC, so, this together with Joel's paper, which apparently not uses AC, answers the question. When I read Joel's comment, I mistakenly thought that Joel's 1998 proof of the full result did not use AC (since this paper of Shelah was written in 2009.) That's why I wanted to double check whether there was some way to avoid Simon's proof back then in Joel's 1998 paper. $\endgroup$
    – Burak
    Jul 17, 2018 at 14:16
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    $\begingroup$ You may also look at Automorphism towers and definability in generalized Baire spaces where details about the Kaplan-shelah's result and related theorems are proved. $\endgroup$ Jul 18, 2018 at 3:00

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