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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?

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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.

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    $\begingroup$ The CKMM theorem is very powerful! Is there a way to read off the group algebra part? $\endgroup$ – Student Apr 19 '19 at 12:24
  • $\begingroup$ 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. $\endgroup$ – Oeyvind Solberg Apr 20 '19 at 7:35
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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 $i \neq j$."

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  • $\begingroup$ Is there a characteristic assumption on the last part? Is there division by 2 in formula for y? $\endgroup$ – AHusain Apr 18 '19 at 22:29
  • $\begingroup$ 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! $\endgroup$ – Student Apr 19 '19 at 0:03
  • $\begingroup$ 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? $\endgroup$ – Student Apr 19 '19 at 0:04
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    $\begingroup$ 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. $\endgroup$ – John Palmieri Apr 19 '19 at 5:28
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    $\begingroup$ @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful. $\endgroup$ – John Palmieri Apr 19 '19 at 5:29
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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.

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  • $\begingroup$ Don't you need algebraic closure of the field for either argument to work? Or am i missing something ? $\endgroup$ – Konstantinos Kanakoglou May 31 '19 at 18:54
  • $\begingroup$ 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 $\endgroup$ – Marco Farinati May 31 '19 at 19:04
  • $\begingroup$ Ok. I was mainly refering to the grading/comodule part but in any case both arguments seem nice and clear. +1 ! $\endgroup$ – Konstantinos Kanakoglou Jun 1 '19 at 1:40
  • $\begingroup$ 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. $\endgroup$ – Marco Farinati Jun 3 '19 at 17:35
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    $\begingroup$ @MarcoFarinati: register and merge your accounts. Then you can edit your own posts. $\endgroup$ – András Bátkai Jun 3 '19 at 18:18

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