Are any subalgebras of the steenrod algebra isomorphic to the group algebra over for some group?

Question

Heading says it all. Wondering if there are any subalgebras of the steenrod algebra which are isomorphic as hopf algebras to $\mathbb{F}_2{G}$ for some group $G$? In particular interest to me are the obvious subalgebras generated by certain $Sq^i$. Is May's book on algebraic steenrod operations a good place to look?

Answer 1

This is too long to be a comment, but it is not a complete answer.

As @QiaochuYuan says, there are no nontrivial grouplike elements, so this cannot happen as Hopf algebras. You can ask instead which Hopf subalgebras of the mod 2 Steenrod algebra $A$ are isomorphic to group algebras as algebras. The starting place for that is the classification of Hopf subalgebras of $A$, originally due to Anderson and Davis, described in various places (here for example).

I don't know of any good "recognition theorem" for group algebras: how to tell if a finite-dimensional $k$-algebra $B$ is isomorphic to $kG$ for some group $G$. So I don't know a complete answer to your question, and I don't know if anyone knows an answer. Some comments:

• the classification of Hopf subalgebras of $A$ includes a classification of those which are isomorphic (as algebras) to group algebras of elementary abelian 2-groups: the Hopf subalgebras which are exterior algebras. See the cited reference.

• If $A(n)$ is the Hopf subalgebra generated by $\mathrm{Sq}^i$ for $i \leq 2^n$, then $A(0)$ is isomorphic to $\mathbb{F}_2 C_2$, and I doubt that any of the others are isomorphic to group algebras. The smallest semidihedral group has order 16, and there is a nice description of their group algebras; they are called "semidihedral algebras." There is actually a semidihedral algebra of dimension 8 — not isomorphic to the group algebra of any group — and it is isomorphic to $A(1)$. (See https://mathoverflow.net/questions/89327/the-semidihedral-group-of-order-16-and-ko and the paper by Crawley-Boevey cited there, for example.)

• The next case would be $A(2)$, which is 64-dimensional. I bet there are invariants of groups (of their cohomology, for example) which will rule out the possibility that $A(2) \cong \mathbb{F}_2 G$ for any group $G$ of order 64. The first cohomology group of $A(2)$ is a 3-dimensional vector space, and that should eliminate most of the possibilities, for instance.