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Let H be a group. Can we find an automorphism $\phi :H\rightarrow H$ which is not an inner automorphism, so that given any inclusion of groups $i:H\rightarrow G$ there is an automorphism $\Phi: G\rightarrow G$ that extends $\phi$, i.e. $\Phi\circ i=i\circ \phi$?

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up vote 25 down vote accepted

The answer is that the inner automorphisms are indeed characterized by the property of the existence of extensions to larger groups containing the original group. I learned as much from this blog entry (in Russian). The reference is Schupp, Paul E., "A characterization of inner automorphisms.", Proc. Amer. Math. Soc. 101 (1987), no. 2, 226--228.

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Where by "The answer is "no"", you are referring to the question in the body of the post, rather tha nthe question in the title. – Theo Johnson-Freyd Dec 4 '09 at 21:42
Oh, so this was ambiguous. I've edited it out. Thanks for pointing this to my attention. – Leonid Positselski Dec 4 '09 at 22:48

I struggled with the same question for quite some time and solved it for finite groups, only to then discover that it had already been solved. Schupp solved it for arbitrary groups. Martin Pettet later solved it for finite groups, and his proof works for p-groups, finite p-groups, finite pi-groups, solvable groups, etc.

  1. On inner automorphisms of finite groups by Martin R. Pettet, Proceedings of the American Mathematical Society, Volume 106,Number 1, Page 87 - 90(May 1989)

  2. Characterizing inner automorphisms of groups by Martin R. Pettet, Archiv der Mathematik, Volume 55,Number 5, Page 422 - 428(Year 1990)

These proofs also show that analogous statements are true if we replace injective embeddings with quotient maps (i.e., the only automorphisms that can be pulled back over all quotient maps are the inner ones).

I also have some notes on this and similar problems here: extensible automorphisms problem.

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