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I am not very familiar with free metabelian groups, so I apologise in advance if this is trivial.

A group $G$ is said to be complete if every automorphism of $G$ is inner. In this case, $\operatorname{Aut}(G) \cong G$ (for centreless groups, this is an equivalent property to being complete). The following classes of groups are known to have complete automorphism groups: f.g. non-abelian free groups [1], f.g. free nilpotent groups of rank $\neq 1,3$ and class $2$ [2], f.g. non-abelian free solvable groups [3].

Hence we have e.g. $\operatorname{Aut}(\operatorname{Aut}(F_2)) \cong \operatorname{Aut}(F_2)$, where $F_2$ is the free group of rank $2$.

Is it known whether every f.g. non-abelian free metabelian group has complete automorphism group?

Note that Bachmuth [4] showed that the automorphism group of a free metabelian group of rank $>2$ is not metabelian (or solvable), and seemed to indicate that the structure for rank $2$ is somewhat tractable. I am curious if anything further has been shown later.

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References

[1] Dyer, Joan L.; Formanek, Edward, Complete automorphism groups, Bull. Am. Math. Soc. 81, 435-437 (1975). ZBL0304.20030.

[2] Dyer, Joan L.; Formanek, Edward, Automorphism sequences of free nilpotent groups of class two, Math. Proc. Camb. Philos. Soc. 79, 271-279 (1976). ZBL0324.20043.

[3] Dyer, Joan L.; Formanek, Edward, Characteristic subgroups and complete automorphism groups, Am. J. Math. 99, 713-753 (1977). ZBL0367.20041.

[4] Bachmuth, S., Automorphisms of free metabelian groups, Trans. Am. Math. Soc. 118, 93-104 (1965). ZBL0131.02101.

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  • $\begingroup$ I guess that "complete" meant either that $G\to Aut(G)$ is surjective, or that it's bijective, but not just that there exists an isomorphism (in particular I don't think the dihedral groups of order $2^{n\ge 3}$ are viewed as complete. In any case, it's not important here since these are groups with trivial center. $\endgroup$
    – YCor
    Commented Aug 23, 2021 at 17:18
  • $\begingroup$ @YCor Ah yes, you're right, I am really only interested in when $\operatorname{Aut}(G) \cong G$ and mislabelled this property, but as you say it's equivalent here. I've updated the question to clarify this. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 23, 2021 at 17:39
  • $\begingroup$ There are several much more recent papers about automorphism groups of metabelian groups. $\endgroup$
    – markvs
    Commented Aug 23, 2021 at 22:17
  • $\begingroup$ @MarkSapir Did you have any particular in mind? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 24, 2021 at 10:35
  • $\begingroup$ For your question: see, for example, Roman'kov, V. A. Primitive elements and automorphisms of the free metabelian group of rank 3. Sib. Èlektron. Mat. Izv. 17 (2020), 61–76, $\endgroup$
    – markvs
    Commented Aug 24, 2021 at 13:01

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