Every finite abelian group $A$ is isomorphic to its dual group $A^*:=\operatorname{Hom}(A,\mathbb{C}^\times)$. The isomorphism of $A$ with $A^*$ is non-canonical, and one way to make this precise is to say that the functor $A\mapsto A^*$ is contravariant, so this functor cannot be naturally isomorphic to the (covariant) identity functor. I wonder if there is an analogous construction that works for all finite groups. Specifically:
Does there exist a contravariant functor $F$ from the category of finite groups to itself, such that $F(G)$ is isomorphic to $G$ for every group $G$?
The imprecise question I have in mind is: for an arbitrary finite $G$, can one construct a group $G'$ that is non-canonically isomorphic to $G$? The existence of a contravariant functor $F$ as above would be one precise way to answer the imprecise question.
from Hot Weekly Questions - Mathematics Stack Exchange
Julian Rosen
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