Let me start with three examples to illustrate my question (probably vague; I apologize in advance).

  1. $\mathbf{Man}$, the category of closed (compact without boundary) topological manifold. For any $M, N\in \mathbf{Man}$ there is the following theorem (Whittaker) which says that

$$\mathrm{Homeo}(M)\cong_{\text{as groups}} \mathrm{Homeo}(N) \, \textit{ if and only if } \, M\cong_{\text{as manifold}} N$$

  1. $\mathbf{NFields}$, the category of Number fields. For any $F, K\in \mathbf{NFields}$ there is a theorem (Neukirch-Uchida) which says that

$$\mathrm{Gal}(\overline{K}/K)\cong_{\text{as progroups}} \mathrm{Gal}(\overline{F}/F) \textit{ if and only if } K\cong_{\text{as fields}} F $$

  1. $\mathbf{Vect}$, the category of finite dimensional vector spaces. For any $V, W\in \mathbf{Vect}$ we have that

$$\mathrm{GL}(V)\cong \mathrm{GL}(W) \textit{ if and only if } V\cong W $$

In the first and the third examples we see that the automorphism group of an object determines the object. The second example seems to be similar in some sense however it does not admit the same naive interpretation.

$\textbf{Question:}$ What are other non-trivial examples of interesting categories where the automorphism group of an object determines the object itself? Is there a name for such categories? Is there a way to compare and characterize these kind of categories?

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    $\begingroup$ An interesting question. I wasn't aware of the result 1.; is there an easily accessible reference for that? $\endgroup$
    – Todd Trimble
    Jan 16, 2017 at 12:39
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    $\begingroup$ @ToddTrimble I think this paper is the original one Whittaker, James V. On isomorphic groups and homeomorphic spaces. Ann. of Math. (2) 78 1963 74–91 dl.dropboxusercontent.com/u/5188175/whit.pdf $\endgroup$ Jan 16, 2017 at 13:01
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    $\begingroup$ @IgorRivin Thanks! We might as well link to it: mathoverflow.net/questions/92422/… $\endgroup$
    – Todd Trimble
    Jan 16, 2017 at 13:27
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    $\begingroup$ Incidentally, with regard to 2., a little bird told me that the result extends to all global fields (so number fields, and also function fields for curves over a finite field -- the extension is apparently due to Pop). This might be a nicer formulation. $\endgroup$
    – Todd Trimble
    Jan 16, 2017 at 13:30
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    $\begingroup$ On suggestion from another user, let's make this a big list CW. $\endgroup$
    – Todd Trimble
    Jan 16, 2017 at 18:04

5 Answers 5


The following result holds.


(1) $\,$ (Baer-Kaplanski) $\,$ If $G$ and $H$ are torsion groups with isomorphic endomorphism rings $\mathrm{End}(G)$ and $\mathrm{End}(H)$, then $G$ and $H$ are isomorphic, and any ring isomorphism $\psi \colon \mathrm{End}(G) \to \mathrm{End}(H)$ is induced by some group isomorphism $\varphi \colon G \to H$.

(2) $\,$ (Leptin-Liebert) If $G$, $H$ are abelian $p$-groups $(p >3)$ and $\mathrm{Aut}(G)$ is isomorphic to $\mathrm{Aut}(H)$, then $G$ is isomorphic to $H$.

See A. V. Mikhalev, G. Pilz: The Concise Handbook of Algebra, p.74 and the references given therein.

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    $\begingroup$ The first item is about endos not autos. Are the groups abelian? $\endgroup$ Jan 16, 2017 at 15:44
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    $\begingroup$ Yes, I know :-) Regarding the groups: well, the statement of the Baer-Kaplanski theorem in the reference that I posted does not say "abelian", but other references actually seem to require this assumption. Let us say "abelian" for the moment, I will try to look at the original reference by Baer or Kaplanski. $\endgroup$ Jan 16, 2017 at 18:41
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    $\begingroup$ @FrancescoPolizzi $End(G)$ is usually not a ring when $G$ is not abelian. $\endgroup$
    – YCor
    Jan 16, 2017 at 21:53
  • $\begingroup$ @YCor: yes, I was thinking about this. Probably the assumption "abelian" for some reason was tacitly assumed in the statement I read, and unfortunately I still do not have access to the original papers. $\endgroup$ Jan 16, 2017 at 22:45

If $M$ and $N$ are smooth manifolds, then their diffeomorphisms groups are isomorphic if and only if $M$ and $N$ are diffeomorphic. This smooth counterpart to Whittaker's theorem has been proved by Filipkiewicz (Ergodic Theory and Dynamical Systems, 1982).


If $M,N$ are two countable, $\omega$-categorical and $\omega$-stable structures, and $\operatorname{Aut}(M)\cong \operatorname{Aut}(N)$ (as topological groups), then $M$ and $N$ are bi-interpretable.

More generally, $M$ and $N$ only need to be countable, $\omega$-categorical and satisfy the so-called small index property. As far as I can recall, an analogous result is true for metric structures (in continuous logic).


Under the Generalized Continuum Hypothesis, $$2^{\aleph_\alpha}=\aleph_{\alpha+1}\quad(\forall\alpha),$$ sets with no structure (so automorphisms are just bijections) is an example.

Namely, by

Cardinality of the permutations of an infinite set

we have $\text{card Aut}(X)=2^{\text{card}(X)}$, and $$2^{\aleph_\alpha}=2^{\aleph_\beta}\implies\aleph_{\alpha+1}=\aleph_{\beta+1}\implies\alpha+1=\beta+1\implies\alpha=\beta.$$

(I suppose it's easy to check that we need more than ZFC but do not need GCH here.)

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    $\begingroup$ There was on MO a question about the "injectivity of the continuum function". If I remember correctly it was stated there that it is indeed weaker than GCH $\endgroup$ Jan 17, 2017 at 6:42
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    $\begingroup$ mathoverflow.net/questions/165549/… $\endgroup$ Jan 17, 2017 at 6:47
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    $\begingroup$ (modulo the silly objection that the empty set and the one-element set have the same automorphisms) $\endgroup$ Jan 17, 2017 at 8:46
  • $\begingroup$ @GabrielC.Drummond-Cole maybe we can fix that by requiring isomorphic group actions? $\endgroup$ Jan 24, 2017 at 7:10
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    $\begingroup$ While comparing automorphism groups, if you consider group isomorphisms instead of mere bijections, then you do not even need GCH or something beyond ZFC for this to hold. Symmetric groups on different infinite cardinals are not isomorphic. See this post mathoverflow.net/questions/12943/… $\endgroup$
    – Burak
    Oct 7, 2020 at 17:04

Geometric Complexity Theory:

Theorem (Mulmuley and Sohoni [MS]) The permanent (respectively the determinant) polynomial is characterized by its symmetry group.

That is if $P$ is a homogeneous polynomial of degree $m$ in $m^2$ variables and its symmetry group $G_P$ also fixes the permanent (respectively the determinant), then $P$ must be a scalar multiple of the permanent (respectively the determinant).

Landsberg and Ressayre [LR] made progress on Valiant's version of P vs NP using this result.

  • $\begingroup$ The 2nd link doesn't work. $\endgroup$
    – HeinrichD
    Mar 10, 2017 at 11:16
  • $\begingroup$ I can't access the paper, but your gloss seems stronger than the statement of the theorem. Can one conclude that $P$ is a scalar multiple of the permanent / determinant even if we only know its symmetry group fixes the permanent / determinant (hence is contained in the symmetry group of the latter), or must the symmetry group of $P$ equal the symmetry group of the permanent / determinant? (Of course, the strong form in your gloss would imply that these two conditions are equivalent, which maybe is a priori obvious anyway—but not to me!) $\endgroup$
    – LSpice
    Mar 21, 2017 at 2:12
  • $\begingroup$ This is certainly in the same spirit, but not quite the same, as in principle there could be another homogeneous poly of degree m in m^2 variables with a stabilizer group abstractly isomorphic to $G_P$. (Indeed, any poly equivalent to perm satisfies this, but that's rather trivial. But I don't know how to rule out other polys with abstractly isomorphic stabilizer.) Also, such polys are very special, whereas I thought an important part of the question was about whole categories. $\endgroup$ Mar 21, 2017 at 2:12
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    $\begingroup$ @L Spice: containment is enough in these cases. $\endgroup$ Mar 21, 2017 at 2:13

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