# Can a group have a subset that is stable under all automorphisms, but not under inverse?

The title really says it all. For any group $G$ the map $g\mapsto g^{-1}$ gives a (canonical) isomorphism from $G$ to the opposite group $G^\mathrm{op}$; this marks a difference with for instance non-commutative rings where the two might not be isomorphic. As a consequence there is for instance no fundamental difference between studying left and right group actions, even for a fixed group$~G$.

So while groups cannot be in any manner "naturally left-handed", whatever that might be construed to mean, there is still the potential possibility of being able to point to some "chiral" subset of the group, a subset that might have some relation to the group without being in the same relation to the opposite group. It is quite evident that notions such as normaliser or centraliser applied to some (set of) elements will never yield such subset; indeed anything that is a subgroup will not do because it is closed under inverse.

Which brought me to my question:

Can a group $G$ have a subset closed under the action of $\operatorname{Aut}(G)$, but not under the map $g\mapsto g^{-1}$?

Since one has in particular the inner automorphisms, one need only consider unions of conjugacy classes.

I can see no reason why this should be impossible, but it is also hard to construct an example.

• For Abelian groups $g\mapsto g^{-1}$ is an automorphism, so no luck here
• On the other hand most symmetric groups have trivial outer automorphism groups; however here every $g$ is conjugate to $g^{-1}$, no luck either
• In alternating groups a conjugacy class of $S_n$ might fall apart, an an element might not be conjugate to its inverse; the two conjugacy classes would still however be related by an outer automorphism (coming from $S_n$)
• In $GL(n,K)$ most elements are not conjugate to their inverse; however there is an obvious out automorphism of transpose-inverse, and every invertible matrix is similar to its transpose (due to the theory of f.g. modules over a PID, leading to classification of similarity by invariant factors).
• Most "natural" subgroups of $GL(n,K)$ would suffer from the same "outer automorphism" phenomenon as alternating groups do, so they do not look promising.

That's where I am now; I unfortunately do not know a lot of examples where I know enough to easily see whether I can find such subsets. Maybe a good candidate is the general linear group over a skew field (division ring), since the result of similarity to one's transpose does not hold there, as far as I know. But I don't know enough about automorphisms of these groups to known whether they are good candidates.

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There are many examples in the literature of finite $p$-groups of nilpotence class $2$ in which all automorphisms are central, that is, for every automorphism $\alpha$ and every $x \in G$, one has $\alpha(x) = x z$ for some $z \in Z(G)$.

If $G$ is such a group, with $p > 2$, and $g \in G \setminus Z(G)$, then there is no automorphism taking $g$ to $g^{-1}$, otherwise $g^{-1} = g z$ for some $z \in Z(G)$, and $g^{2} \in Z(G)$, which forces $g \in Z(G)$ as $p > 2$.

So you may take the orbit of any such $g$ under the automorphisms of $G$ as your subset.

As a reference, you may take this paper.

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Thank you, this is very informative. I gather that (correct me if I'm wrong) (1) indeed this is indeed possible, even if one requires $G$ to be a finite ($p$-)group, but (2) indeed giving a concrete example is not so easy. Since you clearly studied this matter much more deeply than I did (or have time to), might I ask what size one might expect $G$ to have in a minimal example (of the type you indicated)? Concretely, do you think $|G|=3^3=27$ could work, or should one expect to need to go much further? – Marc van Leeuwen Jun 15 '14 at 8:34
@MarcvanLeeuwen, you're welcome, will reply later. – Andreas Caranti Jun 15 '14 at 9:04
@MarcvanLeeuwen, there are results that say, roughly speaking, that most $p$-groups of class $2$ will have all of their automorphisms central. So finding examples is not too difficult. This is usually done by constructing groups via generators and relations, with the relations meant to break all symmetries. Nothing deep, but some involved calculations can be in order. A paper of Marta Morigi, quoted as [Mor94] in the paper referred to in my answer, shows that you have to go up to $p^{7}$ to find such an example. – Andreas Caranti Jun 15 '14 at 9:57
@MarcvanLeeuwen, following up you comment that "giving a concrete example is not so easy", I am writing up a class of examples that do not require any involved computations, and are easily described in terms of linear algebra. Will notify you when the little manuscript is ready. – Andreas Caranti Jun 19 '14 at 11:22
Great, thanks in advance! By the way I realise that what I was initially looking for, a characteristic subset of a group that might be different for its opposite group (something like a "left radical", whatever that might be) is not what I ended up asking for, but having an answer to the question intrigues me nonetheless. – Marc van Leeuwen Jun 19 '14 at 11:46