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?
