# For which categories we can solve $\text{Aut}(X) \cong G$ for every group $G$?

It is usually said that groups can (or should) be thought of as "symmetries of things". The reason is that the "things" which we study in mathematics usually form a category and for every object $X$ of a (locally small) category $\mathcal{C}$, the set of automorphisms (symmetries) of $X$, denoted by $\text{Aut}_{\mathcal{C}}(X)$, forms a group.

My question is: Which categories that occur naturally in mathematics admit all kinds of symmetries? More precisely, for which categories we can solve the equation (of course up to isomorphism) $$\text{Aut}_{\mathcal{C}}(X) = G$$ for every group $G$?

• Negative for $\mathsf{Set}$: Infinite sets have infinite symmetry groups and for finite sets we get $S_n$'s. So if we let $G$ to be any finite group which is not isomorphic to some $S_n$, the equation has no solution.

• Negative for $\mathsf{Grp}$: No group can have its automorphism group a cyclic group of odd order.
• Positive for $\mathsf{Grph}$ (category of graphs): Frucht's theorem settles this for finite groups. Also according to the wikipedia page, the general situation was solved independently by de Groot and Sabidussi.

• An obvious necessary condition is that $\mathcal{C}$ should be a large category.
• This paper shows that the equation can be solved if $\mathcal{C}$ is the category of Riemann surfaces with holomorphic mappings and $G$ is countable.

• If we take $\mathcal{C}$ to be the category of fields with zero characteristic, I guess the equation relates to the inverse Galois problem. Edit: This may be much easier than the inverse Galois problem, as Martin Brandenburg commented.
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Notice that it works too for edge-labeled directed graphs thanks to Cayley graphs. –  Seirios Jul 2 '12 at 14:00
Community wiki? –  Rasmus Jul 2 '12 at 14:55
Finite posets yield all finite groups, but I'm not sure about infinite groups via infinite posets. –  Thomas Andrews Jul 2 '12 at 16:25
In Set the singleton object 1 is its own group, since $\hom(\mathbf{1},\mathbf{1})\cong\mathbf{1}$...although you may reject this because it's a (canonical) isomorphism, and not an equality... –  Alex Nelson Jul 2 '12 at 16:50
Nice question! But if you want to have any connection with the inverse Galois theory, you should restrict yourself to finite groups and the category of finite Galois extensions of $\mathbb{Q}$. I suspect that the problem for the category of all extensions of $\mathbb{Q}$ is much simpler (perhaps already solved). –  Martin Brandenburg Jul 3 '12 at 16:29

Since the category of finite posets is isomorphic(!) to the category of finite $T_0$ spaces, in particular every finite group is the automorphism group of a topological space. More generally, it is proven in Spaces with given homeomorphism groups by Thornton, also available online, that every finitely generated group is the automorphism group of a topological space. I wonder if we can realize every group?