# Subalgebra of a group algebra

Let $$k$$ be a field, $$G$$ a finite group, and $$k[G]$$ the group algebra. Let $$A$$ be a subalgebra of $$k[G]$$. In general, $$A$$ is not the group algebra of some subgroup $$H$$ of $$G$$.

Question: Is there any criterion for when $$A = k[H]$$ for some subgroup $$H$$? Also, in that case, how do we read of the generating subgroup $$H$$? Will the situation become better/easier if I assume $$A$$ to be a sub-Hopf-algebra?

## 2 Answers

The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $$H$$ over an algebraically closed field $$k$$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $$H$$ is finite dimensional cocommutative Hopf algebra over $$k$$, then $$H$$ is isomorphic to a group algebra.

This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $$p$$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.

See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.

• The CKMM theorem is very powerful! Is there a way to read off the group algebra part? – Student Apr 19 at 12:24
• I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional. – Oeyvind Solberg Apr 20 at 7:35

If $$A$$ is the group algebra of a subgroup, then $$k[G]$$ will be free as a module over $$A$$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.

Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $$x$$ which satisfy $$\Delta(x) = x \otimes x$$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.

Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $$\{y\}$$ satisfying $$y_i^2=y_i$$ and $$y_iy_j=0$$ for $$i \neq j$$."

• Is there a characteristic assumption on the last part? Is there division by 2 in formula for y? – AHusain Apr 18 at 22:29
• Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra! – Student Apr 19 at 0:03
• For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra? – Student Apr 19 at 0:04
• Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual. – John Palmieri Apr 19 at 5:28
• @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful. – John Palmieri Apr 19 at 5:29