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 $\sum_{a\in dA}f(a)$.


  • 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 $\det A:=|\det M|$.


  • 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.


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*}

  • 1
    $\begingroup$ 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. $\endgroup$ Commented May 31, 2019 at 15:15
  • $\begingroup$ @ancientmathematician See my edit $\endgroup$
    – Mastrem
    Commented May 31, 2019 at 16:44
  • 2
    $\begingroup$ @ancientmathematician There's a TeX error somewhere in that formula. Unfortunately, I can't read it. Did you mean to write $$\det M(\prod\mathbb{Z}_{p^{r_j}})=\left(\prod p^{r_j}-\prod p^{r_j -1}\right)\det M(\prod\mathbb{Z}_{p^{r_j -1}})$$? $\endgroup$
    – Mastrem
    Commented May 31, 2019 at 17:00
  • 1
    $\begingroup$ 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 $\endgroup$ Commented May 31, 2019 at 17:02

1 Answer 1


Here's the answer to the first question:

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 $\det A > 0$.


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