Is there a non-trivial group G isomorphic to Aut(G)?

Question

The title basically says it all.

Is there a group with more than one element that is isomorphic to the group of automorphisms of itself?

I'm mainly interested in the case for finite groups, although the answer for infinite groups would still be somewhat interesting.

See http://mathoverflow.net/questions/5635/does-autaut-autg-stabilize — Mariano Suárez-Álvarez, Jul 02 '10 at 00:45
Answer: yes. Smallest example: $S_3$, with order 6. Arguably most interesting example: monster simple group. — S. Carnahan, Jul 02 '10 at 00:55
From Mariano's link is the example of $D_8$, which is isomorphic to its automorphism group, but not by the natural map. — Ben Wieland, Jul 02 '10 at 01:34

José Hdz. Stgo. · Accepted Answer · 2010-07-02T01:41:28.840

29

The automorphism group of the symmetric group $S_n$ is (isomorphic to) $S_n$ when $n$ is different from $2$ or $6$. In fact, if $G$ is a complete group you can ascertain that $G \simeq \mathrm{Aut}(G)$. The reverse implication needn't hold, though.

edited Jul 02 '10 at 01:41

answered Jul 02 '10 at 00:47

José Hdz. Stgo.

8,732

8

As mentioned in the comments above, $D_8$ is a counterexample to the reverse implication, since it is not complete but is isomorphic to its automorphism group. – Joel David Hamkins Jul 02 '10 at 02:02
3

Could you perhaps add the definition of a complete group? – user717 Jul 02 '10 at 09:28
10

Complete = centerless + all automorphisms are inner. Such a group G is isomorphic to Aut(G) by the map taking an element g to the inner automorphism conjugation-by-g. – Joel David Hamkins Jul 02 '10 at 16:26

score 16 · Answer 2 · answered Sep 13 '15 at 22:14

16

I am slightly surprised that Wielandt's automorphism tower theorem has not been mentioned: this asserts that given a finite group $G$ with trivial center, the sequence of groups defined by $G_{0} = G$ and $G_{n+1} = {\rm Aut}(G_{n})$ for $n \geq 0$ eventually stabilizes, that is at some stage ${\rm Aut}(G_{n}) = G_{n}$.

answered Sep 13 '15 at 22:14

Geoff Robinson

42,683

3

This result also extends to infinite centerless groups. (http://math.univ-lyon1.fr/~altinel/Master/M1Logique/simonthomasautomorphismtower.pdf) – Burak Sep 13 '15 at 22:43
@Burak: Thanks for pointing that out. I was not sure about the infinite case. – Geoff Robinson Sep 13 '15 at 22:46
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Prof. Robinson: There is even a MO thread on that pearl: http://mathoverflow.net/questions/5635/does-mathrmaut-mathrmaut-mathrmautg-stabilize – José Hdz. Stgo. Sep 13 '15 at 23:12

score 3 · Answer 3 · answered Sep 14 '15 at 09:55

3

For the absolute Galois group of $\mathbb Q$, the map $g\mapsto (h \mapsto ghg^{-1})$ is an isomorphism between $G_\mathbb Q$ and its automorphism group. This is a corollary of the Neukirch-Uchida theorem.

answered Sep 14 '15 at 09:55

user00000

376
2
7

score 3 · Answer 4 · edited Aug 10 '21 at 22:55

3

See also Problem 2015-4C in the Nieuw Archief voor Wiskunde. A solution (of mine) can be found here: http://www.nieuwarchief.nl/home/problems/pdf/uitwerking-2015-4.pdf

edited Aug 10 '21 at 22:55

José Hdz. Stgo.

8,732

answered Aug 10 '21 at 21:55

Alex Heinis

31

Is there a non-trivial group G isomorphic to Aut(G)?

4 Answers4