According to WP article on Hall algebras one counts the number of abelian subgroups in abelian group with fixed type of subgroup, group, quotient.

Two things are claimed:

1) These numbers are polynomials in "p"

2) Using these numbers one naturally defines the algebra structure on the isomorphism classes of abelian groups which appears to be associative and commutative.

Question: What happens if we consider all p-groups, not just abelian one ? Will same/similar claims be true ?


The natural context for the question seems to me some categories with finite number of exact triples A->B->C for fixed A,B,C. So natural generalization - what the properties of the categories for which we can define associative algebra ? commutative algebra ?


Well, the first claim would be impossible to properly describe in the way you want, because unlike abelian $p$-groups which are always indexed by partitions, non-abelian $p$-groups don't have the same classification for different $p$: you don't even necessarily have the same number of groups of order $p^n$ for fixed $n$ and varying $p$!

Regarding the second claim, for any given $p$ the algebra generated should still be associative, as one can still count filtrations with the desired quotients, but commutativity will be lost; for instance, given $p=2$, the quaternion group $Q_8$ can be generated from a short exact sequence

$$1 \rightarrow \mathbf{Z}/2\mathbf{Z} \rightarrow Q_8 \rightarrow \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} \rightarrow 1 $$

but there exists no injective map $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z} \rightarrow Q_8 $, so the products of the elements corresponding to those two groups do not equal each other.

The proper generalization is instead to consider abelian categories where $\hom(M,N)$ and $\mathrm{Ext}^1(M,N)$ are finite. Aside from the classical case of abelian $p$-groups, I believe the best studied is that of quiver representations over a finite field $\mathbf{F}_q$; these, too, form polynomials over $q$, though are non-commutative in all but trivial cases.


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.