I was wondering if it is possible to classify the finite groups which have no outer automorphisms?

I am currently only aware of the Symmetric Groups ($n \neq 6$) as an infinite class of examples. If there is no classification I would still appreciate any further example of groups which have no outer automorphisms.

By asking GAP to go through the groups of order (up to 110) I found the following groups with trivial outer automorphism:

1, S2, S3, C5 : C4, S4, C7 : C6, C9 : C6, C11 : C10

(These are how GAP describes these groups when I used the function "StructureDescription" and I believe : indicates some sort of semidirect product. In these cases I think the smaller cyclic group may be isomorphic to the automorphism group of the larger one and the action is the action of automorphisms.) From this data, it looks as though there could be a class of groups $C_n \rtimes Aut(C_n)$ which have no outer automorphisms but I am unsure for exactly which $n$ this will be the case.

[Edit: ran through groups of order 1 .. 255 on GAP: 1, C2, S3, C5 : C4, S4, C7 : C6, C9 : C6, S5, S3 x (C5 : C4), (C3 x C3) : QD16, S3 x S4, C13 : C12, (C2 x C2 x C2) : (C7 : C3), (C2 x C2) : (C9 : C6), S3 x (C7 : C6)]

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    $\begingroup$ All automorphism groups of nonabelian simple groups are complete (which means trivial centre and no outer automorphisms). It is unlikely that there could be a complete classification because there are too many diverse examples. There are some examples of odd order for instance. Yes, $C_p \rtimes {\rm Aut}(C_p)$ is certainly complete for $p$ prime. $\endgroup$ – Derek Holt Nov 6 '18 at 15:01
  • $\begingroup$ @Derek Holt: Doesn't $\mathfrak{A}_n$ have outer automorphisms? $\endgroup$ – abx Nov 6 '18 at 15:34
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    $\begingroup$ @abx: Derek Holt stated that for a nonabelian finite simple group $G$, the group $\mathrm{Aut}(G)$ is complete. Such a group $G$ is not itself complete, except for rare exceptions (e.g. groups of Lie type over $\mathbb{F}_{p^r}$ with $r > 1$ have the frobenius as an outer automoprhism). $\endgroup$ – js21 Nov 6 '18 at 15:55
  • $\begingroup$ Ah, OK, sorry I read too fast. $\endgroup$ – abx Nov 6 '18 at 17:34

Wielandt's automorphism tower theorem states that if $G$ is any finite group with $Z(G) = 1$, then the sequence of groups $G_{n}$ with $G_{0}= G$ and $G_{n+1} = {\rm Aut}(G_{n})$ is eventually stable (and it follows that the sequence stabilizes with a complete group- ie a group without outer automorphisms).

Later edit, rewording after @YCor's comment. Hence each finite group with trivial center gives rise to a uniquely determined finite group with trivial center AND with trivial outer automorphism group.

  • $\begingroup$ Informally, you get a natural "retraction" of the "set" of finite groups with trivial center onto its "subset" of complete finite groups with trivial center. I don't see how you derive the latter consequence. $\endgroup$ – YCor Nov 6 '18 at 16:45
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    $\begingroup$ Yes, I agree. But that every element in a class P uniquely determines an element in the class Q does not imply that a classification of groups in Q would give a classification of groups in P. (Stupid example: every group G, say of order $>6$ canonically defines the group $S_{|G|}$ of permutations of $G$, which is complete with trivial center, and we can reasonably classify symmetric groups.) However, it turns out that this tower construction provides a wealth of complete groups: it embeds every group with trivial center as a subnormal subgroup of a complete group with trivial center. $\endgroup$ – YCor Nov 6 '18 at 18:19
  • $\begingroup$ @YCor: Yes, I amended the wording, and the wording is correct as it stands in the amended version ( I had pointed out the conclusion of the last sentence of your second comment after the amendment in response to your first comment. I agree that my wording was loose originally). $\endgroup$ – Geoff Robinson Nov 6 '18 at 18:58

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