# Is every group the automorphism group of a ring?

I know not all groups can be realized as the automorphism group of a group. For example, it is well-known that no group can have $$\mathbb Z/n\mathbb Z$$, with $$n > 1$$ odd, as automorphism group. Now I'm wondering the same question about automorphism groups of rings, is there any result about this?

Since the inverse Galois problem (EDIT: over $$\mathbb Q$$) is widely believed to have an affirmative answer, I guess there is no obstruction for a finite group to be the automorphism group of a field, in particular of a ring, but what about infinite groups?

• Since every group is the automorphism group of a graph (even a cubic graph), can't you just use the Stanley-Reisner ring of the graph over a field of your choice? May 2, 2023 at 19:44
• @DaveBenson I believe that ring will usually have many more automorphisms than those of the graph it comes from. May 2, 2023 at 19:46
• Ah, true. But some construction based on the graph should work... May 2, 2023 at 19:48
• Could you define exactly what you mean by ring, since there are many definitions according to different communities. (Unital? Associative? Commutative?)
– YCor
May 2, 2023 at 19:48
• (Comment: If you accept subgroup of automorphism groups, then it's easy: Every group can be represented on the ring $B(\ell^2(G))$ by left regular representation $g \mapsto U_g$ and is so subgroup of $Aut(B(\ell^2(G)))$ by inner automorphisms induced by $U_g$.) May 3, 2023 at 12:50

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $$G$$ finite

It follows from Artin's theorem that every finite group $$G$$ is the Galois group of some finite Galois extension $$L/L^G$$, but we have no control over $$L^G$$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $$L/\mathbb{Q}$$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $$G$$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $$L/L^G$$ as shown in [Wat74], alas once again we have no control over $$L^G$$.

But once again this is a non-problem in the context of OP's question. Even more generally it is proved in [DugGöb87] that for every prescribed infinite (not just profinite) group $$G$$ and every prescribed base field $$K$$ there exists an extension $$L = L(G,K) / K$$ (in their notation $$R(K,G)$$) such that $$\operatorname{Aut}_K(L) \cong G$$. Taking $$K:=\mathbb{Q}$$, we get $$G \cong \operatorname{Aut}(L)$$ (the full automorphism group of $$L$$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $$K$$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $$G$$ and $$K:=\mathbb{Q}$$, and secondly, we can simply take $$K \cong \bar{K}$$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

• Every finite group is the automorphism group of some finite extension of $\mathbf Q$ (not necessarily Galois!). See ams.org/journals/proc/1980-080-03/S0002-9939-1980-0580989-8/… May 2, 2023 at 21:34
• I think the OP's "no obstruction" comment meant "we expect each finite group $G$ to occur since we expect $G$ to be isomorphic to some ${\rm Gal}(K/\mathbf Q)$" which makes $G \cong {\rm Aut}(K/\mathbf Q)$. I don't think the OP knew that this last condition was already settled without requiring $K$ to be Galois over $\mathbf Q$. May 2, 2023 at 23:10
• The paper's title is misusing the term "Galois group". A Galois extension of a field, among other things, must be algebraic, and an infinite Galois group in that setting can be equipped with a topology (of Krull) making it compact, Hausdorff, and without isolated points. A topological space that is compact, Hausdorff, and without isolated points is uncountable. So it is impossible for an infinite-dimensional Galois extension of a field to have a countable Galois group. Thus a paper saying every infinite group is a Galois group is not using the term Galois group in the standard way (contd...) May 3, 2023 at 0:32
• I would agree that the paper could be showing every infinite group is the automorphism group of some field, but not that it is the Galois group of some field in the sense of Galois theory. The link I gave above in my second comment is to a paper showing each finite group is the automorphism group of some finite extension of $\mathbf Q$ (but no claim is made in the paper about the field extension being Galois over $\mathbf Q$), so together with the paper you mention we could say each group arises as the (full) automorphism group of some field. So the OP's question has answer yes using fields. May 3, 2023 at 0:38
• @AlecRhea perhaps so, but I think one needs to be careful when describing these results about realizing groups as automorphism groups of fields so that it doesn't give the impression that they have done something like settle the inverse Galois problem in the usual meaning of that term. May 3, 2023 at 16:57

Every group, finite or infinite, is the full automorphism group of a field. This is a theorem attributed in a few places to the following citation:

E. Fried and J. Kollar. Automorphism groups of fields. In B. Cs´ak´any, E. Fried, and E.T. Schmidt, editors, Universal Algebra (Proc. Conf. Esztergom 1977).

I can't seem to find a copy of that paper, but I can link a place where it is proved below. I think this is a follow up paper to the Fried-Kollar paper mentioned by M.G. in the other answer (but I can't find it)

The proof is a bit long to recreate here, and I don't know off hand all of the details, but very roughly this is the idea. You begin by constructing a graph with the specified group as it's automorphism group. To build the field, you adjoin a transcendental element to $$\mathbb{Q}$$ for each vertex in the graph, then add more elements to the field to encode the edge relation and remove any algebraic automorphisms.

The fields constructed this way are very far from being galois extensions of $$\mathbb{Q}$$, so the inverse galois problem have very little to do with it.

The classic proof is in section 4.3 of The Automorphism Tower Problem (Simon Thomas). This is where I know the theorem from.

I found This Paper of Kaplan and Shelah while looking for the Fried-Kollar paper. It claims to have a simpler proof, but I have not read it yet.