Let $G$ be an abstract group. If we can embed $G$ into a group $H$, in a way that we had $G$ and $H$ of the same Hopf algebra of representative $k$-valued functions ($R(G)\sim R(H)$ as Hopf algebras). My question is when can we have $G$ isomorphic to $H$ ? What sufficient conditions, if needed, to have this? Thanks
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2$\begingroup$ One can check that the group like elements in $R(G)$ and $R(H)$ are precisely $G$ and $H$. Hence any Hopf algebra isomorphism of $R(G)$ and $R(H)$ factors as a group isomorphism of $G$ and $H$. $\endgroup$– estebanCommented Apr 17, 2023 at 6:58
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$\begingroup$ What is "the Hopf algebra of $k$-valued functions" if $G$ is infinite? $\endgroup$– abxCommented Apr 17, 2023 at 8:06
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1$\begingroup$ I suppose it is understood as the algebra of finitely supported functions. $\endgroup$– estebanCommented Apr 17, 2023 at 8:12
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1$\begingroup$ @Sein: Probably, but then the grouplike elements are the homomorphisms $G\rightarrow k$, which do not suffice to recover $G$. $\endgroup$– abxCommented Apr 17, 2023 at 12:00
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$\begingroup$ Yes i in fact meant the hopf algebra of representative functions. @sein does this deal of searching grouplike elements works for the hopf algebra of representative functions? Or your answer was for the group algebra maybe! Thanks $\endgroup$– user502786Commented Apr 18, 2023 at 1:03
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This question is not well formulated because you don't define $R(G)$. Also it is not clear to me what you want to know,since your "if" clause has no "then" clause.
I assume you mean the Hopf algebra of representative functions not the group algebra since you can recover $G$ as the grouplikes in the group ring.
So $R(G)$ consists of the functions which generate a finite dimensional submodule of $k^G$ under the natural module structure. Your question is essentially answered in Hopf dual of the Hopf dual. There are finitely generated groups with no non-trivial finite dimensional representations. For instance any finitely generated infinite simple group has no non-trivial finite dimensional reps finitely generated linear groups are residually finite by Malcev. So only constant functions are representative.
So you can take Thompson's groups $T$ and $V$. They are infinite simple finitely generated groups and $T$ is contained in $V$. They are not isomorphic. I'm not sure you can find decent sufficient conditions outside of both groups being finite.
answered Apr 17, 2023 at 19:47
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$\begingroup$ Yes i mean hopf algebra of representative functions, if we take in particular the context of tannaka-krein duality, and G a group embedded in the characters group X(R(G)) of R(G), such that R(G) is isomorphic to R(X(R(G)) as hopf algebras, does this implies that G isomorphic to X(R(G)) ? $\endgroup$ Commented Apr 18, 2023 at 1:27
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$\begingroup$ But XR(G) is commutative is it not? So why would G be embedded in XR(G)? $\endgroup$ Commented Apr 18, 2023 at 2:01
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$\begingroup$ no it is not commutative unless G is, as in fact it is a group under the convolution product $\endgroup$ Commented Apr 18, 2023 at 4:03
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$\begingroup$ And G i can suppose that it is embedded in it under some assumption on G. $\endgroup$ Commented Apr 18, 2023 at 4:06
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$\begingroup$ How do you define the convolution of characters so that you get another character? $\endgroup$ Commented Apr 18, 2023 at 10:55