# Proving this Hall algebra is commutative without Matlis duality

For a finite abelian $$p$$-group $$G$$ we have that $$G \simeq \mathbf{Z}/(p)^{\lambda_1} \oplus \dotsb \oplus \mathbf{Z}/(p)^{\lambda_r}$$ for some positive integers $$\lambda_1 \geq \dotsb \geq \lambda_r$$. Note that $$G$$ is uniquely determined by $$p$$ and this partition $$\lambda = (\lambda_1, \dotsc, \lambda_r)$$, so let's call $$\lambda$$ the type of $$G$$. For types $$\lambda$$, $$\mu$$, and $$\nu$$, define the Hall number $$g_{\mu,\nu}^\lambda(p)$$ to be the number of normal subgroups $$N \mathrel{\triangleleft} G$$ of type $$\nu$$ such that $$G/N$$ has type $$\mu$$. These Hall numbers serve as the structure constants of an associative algebra called the Hall algebra.

It turns out that this algebra is commutative, i.e. $$g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$$. The proof of this that I'm looking at, following the more general theory in MacDonald's Symmetric Functions and Hall Polynomials, goes like this: You realize that we're looking at the category of finite-length modules over $$\mathbf{Z}_p$$, the $$p$$-adic integers. The Prüfer $$p$$-group $$\mathbf{Z}(p^\infty)$$ is the injective hull of $$\boldsymbol{k} = \mathbf{Z}/(p)$$ in this category, and the functor $$\mathrm{Hom}({-},\mathbf{Z}(p^\infty))$$, via Matlis duality, gives you a bijection of the short exact sequences in question, so $$g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$$.

Proving this can also be approached by developing the theory of characters of finite abelian groups, section 3 in particular. But this is really the same approach in a different language: $$\mathbf{Z}(p^\infty)$$ plays the role of $$S_1$$ in this context. But in either approach, we're introducing some heavy stuff just to prove a fact about $$p$$-groups and partitions. Is there a elementary way to prove that $$g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$$ in the case of finite abelian $$p$$-groups?

• You can replace $\mathbf{Z}\left(p^\infty\right)$ in the duality argument by its "finite approximation" $\mathbf{Z}/p^N\mathbf{Z}$, where $p^N$ is an upper bound on the sizes of your $p$-groups. This makes everything more elementary (it is certainly a lot easier to prove that each group $\mathbf{Z}/p^k\mathbf{Z}$ is isomorphic to its "$N$-dual" group $\operatorname{Hom}\left(\mathbf{Z}/p^k\mathbf{Z},\mathbf{Z}/p^N\mathbf{Z}\right)$ when $k \leq N$). – darij grinberg Feb 22 at 3:57
• Then we don't have to bring up the Prüfer group at all. :) But then we have to do some manual labor to show the map $\mathrm{Hom}(G, \mathbf{Z}/p^N\mathbf{Z}) \to \mathrm{Hom}(N, \mathbf{Z}/p^N\mathbf{Z})$ is a surjective. And we've still gotta define $\mathbf{Z}_p$ because these are all $\mathbf{Z}_p$ modules. ... or do you? – Mike Pierce Feb 22 at 4:30
• You don't need to define $\mathbf{Z}_p$; you can read "finite abelian $p$-group" for "$\mathbf{Z}_p$-module" (since all of your $\mathbf{Z}_p$-modules are finite). – darij grinberg Feb 22 at 4:52
• Do you mean $G \cong \mathbf{Z}/(p)^{\lambda_1} \oplus \dotsb \oplus \mathbf{Z}/(p^r)^{\lambda_r}$? – Lord Shark the Unknown Feb 22 at 5:41
• Yes, that's equivalent to what he is writing. (We always have $\left(n\right)^k = \left(n^k\right)$ as ideals.) – darij grinberg Feb 22 at 5:50

The commutativity of the Hall algebra is saying that you can 'turn short exact sequences around', so is essentially equivalent to having a duality. You don't necessarily need the Prüfer group, though. You can take the 'usual' injective $$\mathbb Q/\mathbb Z$$ for abelian groups. Then $$\mathrm{Hom}_{\mathbb Z}(\mathbb Z/p^n\mathbb Z,\mathbb Q/\mathbb Z)\cong\mathbb Z/p^n\mathbb Z$$ is clear, sending a homomorphism $$f$$ to the image of the cyclic generator $$f(1)$$.
In the 'algebraic' setting, rather than the 'number theoretic' setting, the same result holds. Here you are taking finite dimensional $$k[t]$$-modules on which $$t$$ acts nilpotently; equivalently finite dimensional $$k[[t]]$$ modules. In this case one can instead use the usual vector space duality $$D=\mathrm{Hom}_k(-,k)$$. When the field $$k$$ is finite, the corresponding Hall algebra is symmetric.