So in the interest of gaining a better understanding of a conjecture (due to Scott, 1960) on the automorphism series (first part of the automorphism tower, no direct limits) of a finite group that every finite group has an automorphism series that is eventually constant (periodic w/ period 1). I used GAP to create a complete list up to order $511$ of those finite groups which satisfy $G\simeq Aut(G)$ (henceforth called $Aut$-stable groups). Studying the list briefly made me notice a number of interesting patterns. One of the most striking features is that all of these groups were either centerless, or had $Z(G)\simeq\mathbb{Z}_2$. These are precisely the two groups which have trivial automorphism group, and I suspect that this fact is true of all (finite) $Aut$-stable groups. The condition is not sufficient on its own (e.g. there are plenty of centerless groups that aren't $Aut$-stable, but thanks to the classic theorem of Wielandt, 1939, every centerless group has an automorphism series that stabilizes in finitely many steps (sidenote: if anyone has access to a detailed English version of the proof, I'd love to see it; original is in German)), so this leaves me to the task of attempting to ascertain what conditions are necessary and sufficient for $Aut$-stability.

So my (somewhat broad) question is: What conditions, for a finite group, are necessary and/or sufficient for $Aut$-stability? I am not necessarily expecting a full answer as this question could very well be open at present, but anything that might help shed some light on the why and how is welcome. (Additionally, if this topic has been studied in the literature, references would be appreciated.)

The list of $Aut$-stable groups of order up to $511$, with GAP ids and structure descriptions:


$(6,1)\simeq S_{3}$

$(8,3)\simeq D_{8}$

$(12,4)\simeq D_{12}$


$(24,12)\simeq S_{4}$



$(48,48)\simeq\mathbb{Z}_{2}\times S_{4}$





$(120,34)\simeq S_{5}$

$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$


$(144,183)\simeq S_{3}\times S_{4}$





$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$

$(252,26)\simeq S_{3}\times(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$




$(324,118)\simeq S_{3}\times(\mathbb{Z}_9\rtimes\mathbb{Z}_3)\rtimes\mathbb{Z}_2)$

$(336,208)\simeq PSL(3,2)\rtimes\mathbb{Z}_2$

$(342,7)\simeq (\mathbb{Z}_{19}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_2$



$(432,520)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{2}$



$(432,734)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$

$(480,1189)\simeq(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})\times S_{4}$





1 Answer 1


Your "centerless" observation is on mark.

If $G$ is a complete group then $G\cong Aut(G)$. A complete group is in particular centerless.

Incidentally, the converse is not true: for instance take the Dihedral group $D_8$.

  • $\begingroup$ $D_8$ appears to be exceptional in that, all other groups in the list I gave that aren't centerless (most of them are) can be expressed as direct products of the form $\mathbb{Z}_2\times C$ where $C$ is a complete group. $\endgroup$ Jan 6, 2017 at 5:29
  • $\begingroup$ I don't think that 'centreless' is a consequence of the abstract isomorphism $G \cong \mathrm{Aut}(G)$—at least, not obviously so—but rather of the *particular* isomorphism $g \mapsto \mathrm{Inn}(g)$. (Of course your statement about complete groups remains true, at least according to the Wikipedia definition, which currently builds 'centreless' into the definition ….) I bring this up because it seemed that @JustinBenfield was asking about abstract isomorphism. $\endgroup$
    – LSpice
    Jul 17, 2018 at 14:16
  • $\begingroup$ Also, in case your counterexample is excessively easy for someone to understand, a more complicated one is $\mathrm S_6$. $\endgroup$
    – LSpice
    Jul 17, 2018 at 14:18

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