Suppose $G$ and $H$ are groups and we have an equivalence of categories between $G\textrm{-}\mathbf{Set}$ and $H\textrm{-}\mathbf{Set}$. (One can think of this as a form of "nonlinear Morita equivalence".) What can be said about $G$ and $H$? I suspect that $G$ and $H$ have to be isomorphic, but I can't prove it. If this is too hard in general, I'm also interested in the case that $G$ and $H$ are assumed to be finite.

Here are some things I've tried:

Since $G$-modules are the abelian group objects in $G\textrm{-}\mathbf{Set}$, we get a Morita equivalence between $\Bbb{Z}[G]$ and $\Bbb{Z}[H]$. So if we tensor with any field $k$, we get a Morita equivalence between $k[G]$ and $k[H]$.
Assume for a moment that $G$ and $H$ are finite. If we take $k = \Bbb{C}$, then this means that $G$ and $H$ have the same number of irreducible representations, so the same number of conjugacy classes. If we take $k=\Bbb{R}, \Bbb{Q}, \overline{\Bbb{F}_p}$, we also get the same number of real, rational and $p$-regular conjugacy classes.

While this approach gives some common properties between $G$ and $H$, it's not possible to conclude that $G$ and $H$ are isomorphic, because there are known examples of non-isomorphic finite groups with isomorphic integral group algebras. (See here) This means that we have to use some of the "nonlinear" information from the category $G\textrm{-}\mathbf{Set}$.

Another thing I considered is that the automorphism group of the forgetful functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$ is isomorphic to $G$, by a simple Yoneda-argument, so if we could somehow reconstruct the forgetful functor just from the category $G\textrm{-}\mathbf{Set}$, this would show that $G$ and $H$ must be isomorphic. I haven't been able to do this, but I reconstructed some other functors: The terminal object in $G\textrm{-}\mathbf{Set}$ is a one-point set with a trivial action, denote this $G$-set by $\{*\}$, we have a natural bijection $\operatorname{Hom}_{G\textrm{-}\mathbf{Set}}(\{*\},X) \cong X^G$, where $X^G$ denotes the set of fixed points under the action of $G$. So we can reconstruct the fixed point functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$. The left adjoint of that functor is the functor $\mathbf{Set} \to G\textrm{-}\mathbf{Set}$ which gives each set a trivial $G$-action. Denote $X$ with a trivial $G$-action by $X_{triv}$. We have $\operatorname{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y_{triv}) \cong \operatorname{Hom}_{\mathbf{Set}}(X/G,Y)$, so the functor which sends each $G$-set to the orbit space $X/G$ is left adjoint to the functor which gives each set a trivial $G$-action, so we can reconstruct the functor $X \mapsto X/G$ from the category $G\textrm{-}\mathbf{Set}$. Not sure if that's helpful.

Maybe it's even possible that $G$ and $H$ don't have to be isomorphic? I'm looking either for a counterexample or a proof.

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    $\begingroup$ Maybe the objects with a free $G$-action ($X^G = \emptyset$) have $G$ as sufficiently close to an initial object to identify the object of $G$ with left multiplication? And then, the forgetful functor is represented by $G$? $\endgroup$ Feb 22, 2018 at 21:21
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    $\begingroup$ Not a full answer - but the center of $G$ is isomorphic to the automorphism group of the functor $\operatorname{id}_{G-\mathbf{Set}}$. $\endgroup$ Feb 22, 2018 at 21:51
  • $\begingroup$ Maybe this old post can help... math.stackexchange.com/questions/1375309/… Beppe. $\endgroup$ Apr 20, 2018 at 12:51
  • $\begingroup$ Maybe this old post can help... math.stackexchange.com/questions/1375309/… $\endgroup$ Apr 20, 2018 at 12:52

3 Answers 3


The empty $G$-set is the initial object.

The coproduct of $G$-sets is the disjoint union.

Call a $G$-set “indecomposable” if it is not the coproduct of two non-empty $G$-sets, so an indecomposable $G$-set is just a transitive one.

An epimorphism of $G$-sets is just a surjective map of $G$-sets.

Up to isomorphism, there is a unique transitive $G$-set with an epimorphism to every other transitive $G$-set, namely the regular $G$-set, whose automorphism group is $G$.

So $G$ can be recovered from the category of $G$-sets.

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    $\begingroup$ Or, to follow up on my comment above - $G$ with left multiplication should be the unique (up to isomorphism) object $X$ with $X^G$ initial but $X$ not initial, and which is not decomposable into a coproduct of two such objects. $\endgroup$ Feb 22, 2018 at 22:13
  • $\begingroup$ For a bit more detail, this blog post has a great write up of what Jeremy Rickard described: qchu.wordpress.com/2013/04/01/… $\endgroup$
    – desiigner
    Oct 8, 2019 at 14:47

You can also invoke the following theorem :

Two categories have equivalent categories of presheaves if and only if they have equivalent Cauchy completion.

Then remark that a group (seen as a category with one object) is Cauchy complete : its only idempotent is the neutral, which is obviously spit.

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    $\begingroup$ More specifically and more concretely, in the case of groups, this recovers $G$ as the full subcategory consisting of the only irreducible projective object. $\endgroup$ Feb 26, 2021 at 17:52

I think you already know this, but for the sake of completeness let me show how to prove the claim using your categorical approach.

For any category $\mathcal C$ equivalent to the category of left $G$-sets for some group $G$ we can reconstruct the forgetful functor up to noncanonical natural isomorphism:

The forgetful functor $F : G\text{-}\mathbf{Set} \to \mathbf{Set}$ is naturally isomorphic to $\mathrm{Hom}_{\mathbf{Set}}(\{*\}, F(-))$. It has a left adjoint given by the free $G$-set, that is the disjoint union $Fr(X) := \coprod_{x \in X} G$, where $G$ denotes the regular $G$-set. This shows, that $F \cong \mathrm{Hom}_{G\text{-}\mathbf{Set}}(G, -)$.

As Rickard explained the regular $G$-set is up to non-unique isomorphism characterized as the object, that admits an epimorphism to every connected nonempty $G$-set. Now take such an object $X$. The functor $\mathrm{Hom}_{\mathcal C}(X,-) : \mathcal C \to \mathbf{Set}$ is after composition with the equivalence $\mathcal C \simeq G\text{-}\mathbf{Set}$ naturally isomorphic to the forgetful functor.

Yoneda gives a canonical isomorphism $$ \mathrm{End}(F) \cong \mathrm{Hom}_{G\text{-}\mathbf{Set}}(G,G) \cong G $$ So altogether we have a non-canonical isomorphism $\mathrm{End}(\mathrm{Hom}_{\mathcal C}(X,-)) \cong G$.


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