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Given a group $G $, the set of automorphisms of $G $ also forms a group, $\rm {Aut}(G) $,with composition as the operation (recall that an automorphism of a group is a bijective endomorphism) .

An inner automorphism is one determined by conjugation by some element $g\in G $. That's we have the automorphism $i_g $ given by $i_g (h)=ghg^{-1}\,\forall h $.

I learned from a comment by @LeeMosher on this site that Burnside once conjectured that any class-preserving automorphism is inner. Does anyone know about any progress on this?

Of course the converse is trivial.

(Btw, the reference here is to conjugacy classes. Of course the class equation gives the sizes of these classes. For example, in the case of an abelian group, the class equation consists in all ones.)

For reference, this would have been early in the $20$-th century.

suckling pig
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  • this looks to be false: https://groupprops.subwiki.org/wiki/Class-preserving_not_implies_inner – hunter Mar 25 '21 at 06:01
  • the example given in the above for finite groups has order 32, and the proof uses the language of group cohomology so it's just within the realm of possibility that burnside didn't know of it,although on the other hand burnside spent his whole career thinking about finite groups and the cohomological proof likely has a non-cohomological translation so i suspect it's more likely that he disproved this conjecture himself – hunter Mar 25 '21 at 06:03
  • Ok and thanks @hunter i do see that it's true for symmetric groups, but, of course other than $n=2,6$ all the automorphism s for $S_n $ are inner... maybe will investigate the groups referred to that arise in differential topology for which it's supposedly also true –  Mar 25 '21 at 06:28
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    It seems Burnside himself found a couterexample: see the last (so far) comment to this answer: https://math.stackexchange.com/a/3838521/870827 –  Mar 25 '21 at 07:45
  • Ahh @rhesu so there we have it. I have thoughts of obtaining that book. but i'll probably just lose it like i did all my other books... just have one now, vinogradov's number theory. sorry for rambling. –  Mar 25 '21 at 07:48
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    @rhesu FYI, the timestamp on a comment is a link to that comment: https://math.stackexchange.com/questions/3838188/inner-automorphism-group-as-the-kernel-of-a-homomorphism/3838521#comment7918217_3838521 – Teepeemm Mar 25 '21 at 13:18

5 Answers5

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It is true for finite simple groups. This is a result of Walter Feit and Gary Seitz (1984). Seitz communicated this to me in a letter and I am not sure if it was formally published afterwards. I needed the result for classifying finite groups whose complex irreducible characters are all primitive, see here (in particular $(6.1)$).

Note (added April 20th 2021): the result of Feit and Seitz appeared in On finite rational groups and related topics, Illinois Jour. of Math., vol. 33 (1989), pages 103-131.

Nicky Hekster
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Unless I have made a mistake, the group 43 of order 32, isomorphic to $$ \langle (2,6)(5,8), (2,8)(3,7)(5,6), (1,2)(3,8)(4,6)(5,7) \rangle $$ is a counterexample of minimal order. The automorphism $$ [ (2,6)(5,8), (2,8)(3,7)(5,6), (1,2)(3,8)(4,6)(5,7) ]\rightarrow [ (1,4)(3,7), (1,4)(2,5)(6,8), (1,2)(3,8)(4,6)(5,7) ] $$ is not inner, but preserves all conjugacy classes.

ahulpke
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This is true for free groups and for surface groups. In fact, in the stunningly beautiful paper*, Edna Grossman proved the following:

Theorem. If $G$ is a finitely generated, conjugacy separable group with every class preserving automorphism inner, then $\operatorname{Out}(G)$ is residually finite.

I mentioned that surface groups satisfy this property, while they are also conjugacy separable. Hence, their outer automorphism groups are residually finite. However, their outer automorphism groups are of particular interest: they are mapping class groups. So we have the following.

Corollary. Mapping class groups of compact orientable surfaces are residually finite.

*Grossman, Edna K. "On the residual finiteness of certain mapping class groups." J. Lond. Math. Soc. (2) 9 (1974/75): 160-164. (doi)

user1729
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Burnside himself answered in 1913$^\dagger$ the question he had posed in 1911: for each prime $p \equiv ±3 \pmod 8$, there is a group $G$ of order $p^6$ and nilpotency class $2$ such that $\operatorname{Out}_c(G)\ne 1$.

$^\dagger$W. Burnside, “On the outer isomorphisms of a group”, Proceedings of the London Mathematical Society, Volume s2-11, Issue 1 (1913), p. 40-42.

Kan't
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I found an answer due to G.E. Wall, a linear group, again of order $32$, so quite probably isomorphic to the answer I accepted. It's problem $3.37 c)$ on page $24$ of John D. Dixon's "Problems in Group Theory".

This example is smaller than the answer Burnside himself provided, $p^6$.

Also, the group is not simple (by @NickyHekster's answer). But that's obvious anyway, because it's a $p$-group.

suckling pig
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    I checked that the example by Wall that you mention is the same as that given by Alex Hulpke (namely SmallGroup(32,43)). But there is also one more group of order $32$ with this property, SmallGroup(32,44). There are many more examples among groups order $64$. – spin Jun 23 '22 at 12:19
  • Also, these examples are not simple, since $p$-groups of order $> p$ are obviously not simple! – spin Jun 23 '22 at 12:20
  • Right they're not. Thanks @spin – suckling pig Jun 23 '22 at 13:10