# Motivation

Let $$A$$ be some finite abelian group and $$f:A\to\mathbb{Z}_{\ge 0}$$ some function. I'd like to evaluate $$f$$, but I only have the following information:

• For all $$a,b\in A$$, if $$\operatorname{ord}(a)=\operatorname{ord}(b)$$, then $$f(a)=f(b)$$
• For all divisors $$d$$ of $$|A|$$, I know the value of $$\sum_{a\in dA}f(a)$$.

# Definitions

• For any finite group $$G$$ and positive integer $$n$$, define $$\psi_G(n)$$ as the number of elements of $$G$$ of order $$n$$.
• For any finite abelian group $$A$$, let $$n:=|A|$$ and enumerate the divisors $$d\mid n$$ such that there exists at least one $$a\in A$$ with $$\operatorname{ord}(a)=d$$ as $$d_1,\ldots,d_m$$. Let the $$m\times m$$ matrix $$M$$ be given by $$M_{ij}=\psi_{d_iA}(d_j).$$ We define $$\det A:=|\det M|$$.

# Questions

• Can you classify all $$A$$ with $$\det A$$ non-zero?
• In general, is there anything interesting known about $$\det A$$?
• Given that such an $$f$$ exists, can you solve my original problem?

# Partial result 1

If $$A$$ is the cyclic group of order $$n$$, $$C_n$$, then the 'determinant' is non-zero and there is a solution to the original problem.

If $$A=C_n$$ there exist elements of order $$d$$ for each positive divisor $$d$$ of $$n$$, and $$dC_n\cong C_{n/d}$$. On top of this, $$\psi_{C_n}(d)=\phi(d)$$, where $$\phi$$ is the totient function.

In terms of the original problem, this means that there exists a function $$g:\mathbb{Z}_{\ge 1}\to\mathbb{Z}_{\ge 0}$$ with $$f(a)=g(\operatorname{ord}(a))$$ for all $$a\in A=C_n$$, and we have the value of $$\sum_{a\in dC_n}f(a)=\sum_{e\mid \frac nd}g(e)\phi(e)$$ for all divisors $$d$$ of $$n$$. Let's say this value is $$v_{d}$$, then $$g(d)=\sum_{e\mid d}\mu\left(\frac de\right)v_e$$ gives a unique solution to the original problem and this also shows $$\det C_n\neq 0$$.

# Partial result 2

As suggested by ancientmathematician, if $$A$$ and $$B$$ are two finite abelian groups of coprime order, if the right enumeration is chosen, the matrix associated with $$A\times B$$ is the Kronecker product of the matrix associated with $$A$$ and the matrix associated with $$B$$.

Proof: Enumerate the orders of $$A$$ as $$d_i$$ and those of $$B$$ as $$e_j$$. For all $$i,j$$ we have $$(d_i,e_j)=1$$ which means that if $$\operatorname{ord}(a)=d_i$$ and $$\operatorname{ord}(b)=e_j$$ for some $$a\in A$$ and $$b\in B$$, then $$\operatorname{ord}((a,b))=d_ie_j$$. This means that the matrix associated with $$A\times B$$ has a row for each $$d_ie_j$$. Next, for all $$i,j,k,l$$, \begin{align*} \psi_{d_ie_j(A\times B)}(d_ke_l) &= \psi_{d_ie_jA\times d_ie_jB}(d_ke_l)\\ &= \psi_{d_iA\times e_jB}(d_ke_l)\\ &= \psi_{d_iA}(d_k)\cdot \psi_{e_jB}(e_l) \end{align*} Enumerating the orders of $$A\times B$$ as $$d_1e_1,d_1e_2,\ldots,d_2e_1,d_2e_2,\ldots,$$ now gives the desired result.

As a consequence, if $$A$$ and $$B$$ are two finite abelian groups of coprime order, there exist positive integers $$n,m$$ with $$\det A\times B = (\det A)^n(\det B)^m.$$

This means that if $$\det A\neq 0$$ for all abelian groups $$A$$ of prime power order, then $$\det A\neq 0$$ for all abelian groups.

# Computations

I've computed $$\det A$$ for all abelian groups of order at most $$10$$: \begin{align*} \det C_2 &= 1\\ \det C_3 &= 2\\ \det C_4 &= 2\quad \det C_2\times C_2 = 3\\ \det C_5 &= 4\\ \det C_6 &= 4\\ \det C_7 &= 6\\ \det C_8 &=8\ \quad \det C_2\times C_4=4\quad \det C_2\times C_2\times C_2=7\\ \det C_9 &=12\quad \det C_3\times C_3 = 8\\ \det C_{10} &=16 \end{align*}

• Have you tried to prove that if $(k,l)=1$ and $A, B$ are abelian groups of orders $k,l$ then $M(A\times B)=M(A)\otimes M(B)$? This is true for the most trivial case, namely $A=\mathbb{Z}_p$, $B=\mathbb{Z}_q$ with $p,q$ distinct primes. – ancientmathematician May 31 '19 at 15:15
• @ancientmathematician See my edit – Mastrem May 31 '19 at 16:44
• @ancientmathematician There's a TeX error somewhere in that formula. Unfortunately, I can't read it. Did you mean to write $$\det M(\prod\mathbb{Z}_{p^{r_j}})=\left(\prod p^{r_j}-\prod p^{r_j -1}\right)\det M(\prod\mathbb{Z}_{p^{r_j -1}})$$? – Mastrem May 31 '19 at 17:00
• Sutely in the $p$-group case the determinant for $A$ is something like $|A|-|pA|$ times the determinant for $|pA|$? Sorry about my fail to write out the full formula inTeX – ancientmathematician May 31 '19 at 17:02

Let $$p$$ be a prime, $$r_1,\ldots,r_m$$ be integers with $$r_1\ge \ldots\ge r_m$$ and suppose that $$A\cong \prod_jC_{p^{r_j}},$$ then the $$d\mid |A|$$ such that there exist $$a\in A$$ of order $$d$$ are precisely the powers $$p^i$$ for $$0\le i\le r_1$$. For each of these $$i$$, the group $$p^iA$$ does not contain any elements of order $$p^{r_1-i+1}$$ or higher. This means that if the orders are enumerated with $$d_i=p^i$$, the matrix $$M$$ is an upper triangular matrix with on the main diagonal the values $$\psi_{p^iA}(p^{r_1-i})\ge \psi_{p^iC_{p^{r_1}}}(p^{r_1-i})=\phi(p^{r_1-i})\ge 1.$$ This means that $$\det A > 0$$. With partial result $$2$$ from the question itself, we may conclude that for all finite abelian groups $$\det A > 0$$.