5
$\begingroup$

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 $G$ ?

$\endgroup$
5
$\begingroup$

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 $U$.)

$\endgroup$
2
$\begingroup$

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!!

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Greg Kuperberg Oct 20 '10 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.