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Let $k$ be a field and $G$ a finite group. Is every sub-Hopf algebra over $k$ of the group algebra $k[G]$ of the form $k[U]$ for a subgroup $U$ of $G$ ?

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2 Answers 2

Yes: The dual $H^*$ is a quotient algebra of $k[G]$, and the latter is the commutative function algebra $\{f:G \to k\}$. Thus $H = k[U]$ for a set $U$, and closure under multiplication then makes $U$ a group.

The nice thing about finite-dimensional Hopf algebras over a field: You can turn them upside down. Vector space duality is an involution on finite-dimensional Hopf algebras. (But I think that this particular argument still works in the infinite-dimensional case: $k[G]^*$ has a weak-* topology and $H^*$ is a quotient by a closed ideal, so it should still create $U$.)

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Nice argument. Thanks –  Ralph Oct 20 '10 at 11:44

Moreover, finiteness of $G$ is irrelevant as Greg's proof works there as well. BTW, you don't need to go into $H^\ast$ if you are willing to use coalgebras: any subcoalgebra of $k[G]$ is $k[U]$ for some subset of $G$. $U$ must be a subgroup for the subco to be subHopf...

A more interesting question is to ask about forms of $k[G]$, i.e. Hopf algebras over a subfield $m$ such that $k\otimes_m H \cong k[G]$. There are a plenty of those which are not group subalgebras!!

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Forms of $k[G]$ are complementary to the work of Ahmad Chalabi, who studied the behavior of $k[G]$ under field extensions. So, better to look for those than for weapons of mass destruction. –  Greg Kuperberg Oct 20 '10 at 11:26

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