# 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?

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 '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 '19 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 '19 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 '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 '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 '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 '19 at 5:29

Yes, there is a criterium. Assuming $$G$$ finite, $$A$$ is of the form $$k[H]$$ for some subgroup of $$G$$ if and only if $$A$$ is a sub-bialgebra (and since $$G$$ is finite, if and only if is Hopf subalgebra).

Clearly $$k[H]$$ is a Hopf subalgebra of $$k[G]$$, but if you take any subalgebra $$A$$, the fact that $$A$$ is also a subcoalgebra means that $$A$$ is a subcomodule of $$k[G]$$, hence $$G$$-graded, so, the $$G$$-homogeneous components are $$1$$-dimentionals and $$A$$ is generated by group-like elements. But a subcoalgebra of the form $$k[X]$$ with $$X\subseteq G$$ is subalgebra only when $$X$$ is a subgroup.

(By the way, if $$G$$ is infinite everything works fine except that $$H$$ is maybe a submonoid and not a subgroup).

An alternative proof avoiding the grading/comodule argument is the following: $$A$$ being subcoalgebra means $$A^*$$ is an algebra quotient of $$k[G]^*=k^G$$= the algebra of functions from G to k. But it is clear that any quotient of $$k^G$$ identifies with $$k^X$$ with $$X\subseteq G$$. The rest of the argument is the same.

• Don't you need algebraic closure of the field for either argument to work? Or am i missing something ? – Konstantinos Kanakoglou May 31 '19 at 18:54
• No, you don't. The álgebra structure on $k^G$ is just $t\times k\times \cdots\times k$, the only ideales are puting zeros in some coordinates – Marco Farinati May 31 '19 at 19:04
• Ok. I was mainly refering to the grading/comodule part but in any case both arguments seem nice and clear. +1 ! – Konstantinos Kanakoglou Jun 1 '19 at 1:40
• in the grading/comodule case, the subcoalgebra $A$ verifies $(A)_g\subseteq (k[G])_g$ (because it is a graded subobject). But $(k[G])_g=kg$, so, $A_g=0$ or $A_g=kg$, because a sub-vector space of a 1-dimensional vector space is zero or everything. Again, being algebraically closed is not important, the point is that $k$ is a field. – Marco Farinati Jun 3 '19 at 17:35
• @MarcoFarinati: register and merge your accounts. Then you can edit your own posts. – András Bátkai Jun 3 '19 at 18:18