[Paper Review] The number of chains of subgroups of a finite elementary abelian $p$-group

Marius Tărnăuceanu|arXiv (Cornell University)|Jun 27, 2015

Finite Group Theory Research4 references6 citations

TL;DR

This paper provides an explicit formula for the number of rooted chains of subgroups in a finite elementary abelian $p$-group $\mathbb{Z}_p^n$, resolving a gap left in prior work. Using recurrence relations and combinatorial group theory, it derives a closed-form expression involving products of $p$-factorials, enabling exact computation of chain counts for any $n$ and prime $p$. The key contribution is a compact, computable formula for $F(\mathbb{Z}_p^n)$, the number of $G$-rooted chains.

ABSTRACT

In this short note we give a formula for the number of chains of subgroups of a finite elementary abelian $p$-group. This completes our previous work [5].

Motivation & Objective

• To complete the characterization of rooted subgroup chains in finite elementary abelian $p$-groups, extending prior work that only partially solved the problem.
• To provide an explicit, computable formula for $F(\mathbb{Z}_p^n)$, the number of $G$-rooted chains of subgroups in $\mathbb{Z}_p^n$.
• To resolve the counting problem for rooted chains in $\mathbb{Z}_p^n$ by establishing a recurrence and solving it via combinatorial group theory.
• To connect the chain count to fuzzy subgroup theory, as rooted chains bijectively correspond to fuzzy subgroups, enhancing computational tools in fuzzy group theory.

Proposed method

• Derives a recurrence relation for $x_n$, the number of rooted chains starting from the trivial subgroup, based on subgroups of intermediate order $p^k$.
• Uses the known formula for the number of subgroups of order $p^k$ in $\mathbb{Z}_p^n$, denoted $a_{n,p}(k)$, which is expressed as a ratio of $p$-factorials.
• Defines $f(r) = \prod_{s=1}^r (p^s - 1)$ for $r \geq 1$, with $f(0) = 1$, to simplify the expression of subgroup counts.
• Solves the recurrence $x_n = \sum_{k=0}^{n-1} a_{n,p}(k) x_k$ via induction, yielding a nested sum over chains of indices $1 \leq i_1 < \cdots < i_k \leq n-1$.
• Transforms the solution into the final formula for $F(\mathbb{Z}_p^n)$ by doubling $x_n$ (to account for chains not starting at 1) and substituting $a_{n,p}(k) = f(n)/(f(k)f(n-k))$.
• Validates the formula by computing explicit values for $n = 0$ to $4$, confirming consistency with known results and prior work.

Experimental results

Research questions

• RQ1 What is the exact number of $G$-rooted chains of subgroups in a finite elementary abelian $p$-group $\mathbb{Z}_p^n$?
• RQ2 How can the recurrence relation for the number of rooted chains be solved explicitly to yield a closed-form formula?
• RQ3 What is the relationship between the number of rooted chains and the combinatorics of subgroup lattice structures in $\mathbb{Z}_p^n$?
• RQ4 Can the formula be expressed in terms of $p$-factorials to enable efficient computation for arbitrary $n$ and $p$?
• RQ5 How does the derived formula compare with earlier partial results, such as those in [5], and does it generalize known cases?

Key findings

• The number of rooted chains of subgroups in $\mathbb{Z}_p^n$ is given by $F(\mathbb{Z}_p^n) = 2 + 2f(n)\sum_{k=1}^{n-1}\sum_{1\leq i_1 < \cdots < i_k \leq n-1} \frac{1}{f(n-i_k)f(i_k - i_{k-1})\cdots f(i_1)}$, where $f(r) = \prod_{s=1}^r (p^s - 1)$ and $f(0) = 1$.
• For $n=0$, $F(\mathbb{Z}_p^0) = 1$; for $n=1$, $F(\mathbb{Z}_p^1) = 2$; for $n=2$, $F(\mathbb{Z}_p^2) = 2p + 4$; for $n=3$, $F(\mathbb{Z}_p^3) = 2p^3 + 8p^2 + 8p + 8$.
• The formula confirms and extends the result of Corollary 10 in [5], which computed $f_3 = a_{3,p}$ via direct enumeration.
• The number of unrooted chains $D(\mathbb{Z}_p^n)$ and total chains $C(\mathbb{Z}_p^n)$ follow from $F(\mathbb{Z}_p^n)$ via the identities $D(G) = F(G) - 1$ and $C(G) = 2F(G) - 1$.
• The recurrence $x_n = \sum_{k=0}^{n-1} a_{n,p}(k) x_k$ provides a more efficient computational alternative to the inclusion-exclusion approach used in earlier work.
• The formula is computationally verifiable and yields polynomial expressions in $p$ for each $n$, with coefficients that are integers independent of $p$.