Are there any non-conjugation "extendible automorphisms" in the category of finite groups?

Question

Let $\mathbf{Grp}$ be the category of groups. Given a subcategory $\mathscr{G}$ of $\mathbf{Grp}$ and $G\in\mathit{Ob}(\mathscr{G})$, a $\mathscr{G}$-extendible map on $G$ will here mean an assignment $\beta$ of a function (not necessarily playing well with the group structure, let alone being a morphism of $\mathscr{G}$) $\beta_h$ to each $\mathscr{G}$-morphism $h$ with domain $G$ such that

• $\beta_h$ is a function from the codomain of $h$ to itself, and

• for all $\mathscr{G}$-morphisms $f:H_1\rightarrow H_2, h_1:G\rightarrow H_1$, and $h_2:G\rightarrow H_2$ such that $fh_1=h_2$, we have $f\beta_{h_1}=\beta_{h_2}f.$

If additionally $\beta_h$ is always an automorphism of the codomain of $h$, say that $\beta$ is an extendible automorphism.

If we take $\mathscr{G}=\mathbf{Grp}$ then the situation is well-understood: in An inner automorphism ..., Bergman showed that each extendible map in this sense corresponds to a word, and as an easy consequence every extendible automorphism of a group $G$ "is" conjugation by some element of $G$. However, the first part of this proof breaks down when we look at $\mathscr{G}=\mathbf{FinGrp}$ the category of finite groups since we no longer have access to free products. Indeed (following an incorrect attempt on my part) Bergman observed that there is an extendible map in the $\mathbf{FinGrp}$ sense which does not correspond to a word (coming from the way $\widehat{\mathbb{Z}}$ acts coherently on all finite groups at once); however, this doesn't yield an extendible automorphism.

Are there $\mathbf{FinGrp}$-extendible automorphisms on some finite group which are not given by conjugation by an element of that group?

Answer 1

Finite-extensible automorphisms of finite groups are inner. The proof is rather non-trivial. For the classes of finite nilpotent and finite soluble groups, this is proved in a paper of Pettet, "On inner automorphisms of finite groups". It can be extended to the class of all finite groups using a theorem of Hartley and Robinson, "On finite complete groups". For how to assemble the proof, see the wiki page https://groupprops.subwiki.org/wiki/Finite-extensible_implies_inner

Edit: What the above shows is that given $G$, there is a single complete group $K$ and an inclusion $G\to K$ with the property that the only automorphisms of $G$ that extend to automorphisms of $K$ are inner ones. The group $K$ in question can be taken to be a semidirect product $P\rtimes G$ where $P$ is a finite $p$-group and $p$ is any given prime that does not divide $|G|$.

This does not quite prove what Noah is asking for here, because what he calls an extendable automorphism is really an extension of the automorphism. So there remains the question of whether the identity automorphism of $G$ extends in a non-trivial way.

If $G$ is trivial then we can prove that this cannot happen as follows. The map $\beta$ assigns to every finite group an automorphism in a way that is compatible with group homomorphisms. So by the above theorem it is inner, and takes every subgroup of a group to itself. Moreover, an inner automorphism of a cyclic group is the identity, so $\beta$ is the identity. However, it does not seem so easy to prove this when $G$ is not trivial, so this is an incomplete answer at this point.

Note that by contrast, in the category of finite abelian groups, the trivial automorphism of the trivial group has $2^{\aleph_0}$ extensions to all finite abelian groups. Namely, for each prime, you have two choices: negate or don't negate.

Second edit: I now realise that for non-trivial groups it is not necessarily true that the identity automorphism has only the identity extension. For example, let $G$ be cyclic of order two. Then there are at least two ways to extend the trivial automorphism of $G$ to a map $\beta$ as in the question. The first is the trivial extension, and the second is conjugation by the image of the non-identity element of $G$. I don't know what the extensions are for general $G$, but it seems like an interesting quesiton.

Answer 2

Not a complete answer. Your definition of an extendible map says that $\beta$ is an endomorphism of the forgetful functor $U$ from the under category $G \downarrow \mathrm{FinGrp}$ to $\mathrm{Set}$ sending $G \to H$ to the underlying set of $H$, and your definition of an extendible automorphism says that $\beta$ is an automorphism of the forgetful functor to $\mathrm{Grp}$ sending $G \to H$ to $H$.

For the corresponding functors on $\mathrm{Grp}$, $U$ is representable by the inclusion $G \to G \ast \mathbb{Z}$ ("polynomials in one variable under $G$"), so by the Yoneda lemma $\operatorname{End}(U)$ is isomorphic to endomorphisms of $G \ast \mathbb{Z}$ as a group under $G$. Such an endomorphism is freely determined by the image of $1 \in \mathbb{Z}$, which is an arbitrary element of $G \ast \mathbb{Z}$ (a "polynomial") of the form

$$f(x) = g_1 x^{n_1} g_2 x^{n_2} \dots ;$$

it evaluates on an element $h \in H$ of a group $G \to H$ under $G$ to

$$f(h) = g_1 h^{n_1} g_2 h^{n_2} \dots \in H$$

and in order for it to be an automorphism it must in particular be a homomorphism, and this can be checked universally in the sense that it's true iff

$$f(xy) = g_1 (xy)^{n_1} \dots = f(x) f(y) = (g_1 x^{n_1} \dots )(g_2 y^{n_2} \dots)$$

in "polynomials in two variables" $G \ast F_2$. Now I guess we appeal to some known facts about reduced words in free products to conclude that this is possible iff everything between $x^{n_1}$ and $y^{n_k}$ (where $y^{n_k}$ corresponds to the end of the word) cancels, which gives $k = 1, n_1 = 1$ and $f(x) = gxg^{-1}$. I think this is how Bergman's argument goes.

For $\mathrm{FinGrp}$, $U$ is no longer representable, but it is still "pro-representable" by the inverse system consisting of all finite quotients of $G \ast \mathbb{Z}$ (see e.g. this blog post), and using this we can show that $\operatorname{End}(U)$ can be identified (as a set) with the profinite completion $\widehat{G \ast \mathbb{Z}}$ of $G \ast \mathbb{Z}$. This is sort of tautologous. I have to admit I don't really understand this thing; I guess it's sort of "power series" to the "polynomials" above. I think it has elements which can be written as infinite words $g_1 x^{n_1} g_2 x^{n_2} \dots$ which converge in the profinite topology (where each $n_i \in \widehat{\mathbb{Z}}$). I am betting the above argument generalizes and we still just get conjugations but I am not familiar enough with how to calculate in profinite completions to really say.