Dtjytyk

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?

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 $mathrm{Hom}_{mathbb Z}(mathbb Z/p^nmathbb Z,mathbb Q/mathbb Z)congmathbb Z/p^nmathbb Z$ 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.