We classify finite cyclic groups based on the sum of the orders of their proper subgroups
Let $G$ be a finite cyclic group of order $n$, and let $x$ be a generator of $G$, so $G = \langle x \rangle$. Then $G \cong Z_n$, and by the Fundamental Theorem of Finite Cyclic Groups, we know that for each positive divisor $d$ of $n$, there is a unique cyclic subgroup of order $d$, namely $\langle x^{n/d} \rangle$. The number of positive divisors of $n$ is denoted by $\tau(n)$. Thus, the number of proper subgroups of $G$ is $\tau(n) - 1$, since we exclude the improper subgroup $G$ itself. We wish to study the sum of the orders of all proper subgroups of $G$. This sum can be expressed as: $$ S(G) = \sum_{\substack{d|n \\ d < n}} d $$ where the sum is taken over all positive divisors $d$ of $n$ that are less than $n$. We will call this the cyclic subgroup sum of $G$, and wish to classify three cases: when $S(G) < n$, when $S(G) = n$, and when $S(G) > n$. The definition is stated below for reference.
If $G$ is a finite cyclic group of order $n$,the cyclic subgroup sum of $G$, is defined to be $$ S(G) = \sum_{\substack{d|n \\ d < n}} d $$ where the sum is taken over all positive divisors $d$ of $n$ that are less than $n$.
Let $G$ be $Z_2$. So $n = 2$, where the only positive divisors are $\{1,2\}$, so $\tau(2)=2$. The cyclic subgroup sum is $S(G) = 1$, which is less than $n=2$.
Let $G$ be $Z_6$. Then $n = 6$, the positive divisors are $\{1,2,3,6\}$, so $\tau(6)=4$. The cyclic subgroup sum is $S(G) = 1+2+3 = 6$, thus $S(G) = |G|$.
Let $G$ be $Z_{18}$. Then $n = 18$, the positive divisors are $\{1,2,3,6,9,18\}$, so $\tau(18)=6$. In this case, the cyclic subgroup sum is $S(G) = 1+2+3+6+9 = 21$, so $S(G) > |G|$.
From the examples above, we see that all three cases are possible. In the next section, we will explore some theorems regarding the cyclic subgroup sum.
If G is a finite cyclic group and its cyclic subgroup sum S(G) = 1 then G is a field.
Suppose $G$ is a finite cyclic group of order $n$ and $S(G) = 1$. We want to show that $G$ is a field. Since $G$ is cyclic, we can write $G \cong Z_n$. The cyclic subgroup sum is given by: $$ S(G) = \sum_{\substack{d|n \\ d < n}} d $$ Since $S(G) = 1$, the only proper divisor of $n$ is $1$. This implies that $n$ has no other divisors except $1$ and $n$ itself, which means that $n$ is prime. Therefore, $G \cong Z_p$ for some prime $p$ and is thus a field. $ \qquad \blacksquare $
Observe the converse of the above theorem is false. If $F$ is a finite field, then $F \cong GF(p^n)$ for some prime $p$ and integer $n \geq 1$ [3]. The cyclic subgroup sum of the additive group of $GF(p^n)$ is given by: $$ S(GF(p^n)) = \sum_{\substack{d|n \\ d < n}} p^d $$ For $n \geq 2$, this sum is always greater than $1$. Thus, the converse does not hold in general. However, the follwing theorem does hold for finite fields.
If $F$ is a finite field of order $p^n$, then the cyclic subgroup sum of its additive group $S((F, +)) = \sum_{i=0}^{n-1} p^i$:
Let $F$ be a finite field of order $p^n$. The additive group of $F$, denoted $(F, +)$, is a finite cyclic group of order $p^n$. The positive divisors of $p^n$ are of the form $p^k$ for $k = 0, 1, 2, \ldots, n$. Thus, the proper divisors of $p^n$ are $p^0, p^1, \ldots, p^{n-1}$. The cyclic subgroup sum of $(F, +)$ is given by: $$ S((F, +)) = \sum_{\substack{d|p^n \\ d < p^n}} d = \sum_{k=0}^{n-1} p^k $$ as required. $ \qquad \blacksquare $
We now consider some properties of the cyclic subgroup sum when $G$ is a cyclic group of order $n$ where $n$ is composite. First, we note that if $n$ is composite, then $S(G) > 1$, since there will be proper divisors greater than $1$. Next, we give some definitions.
If $n$ is a positive integer, the aliquot sum of $n$, denoted $s(n)$, is defined to be the sum of all proper divisors of $n$. That is, $$ s(n) = \sum_{\substack{d|n \\ d < n}} d $$
A positive integer $n$ is called perfect if $s(n) = n$, deficient if $s(n) < n$, and abundant if $s(n) > n$.
Notice that the cyclic subgroup sum $S(G)$ of a finite cyclic group $G$ of order $n$ is precisely the aliquot sum $s(n)$ of $n$. Thus, we can classify finite cyclic groups based on whether their order is a perfect, deficient, or abundant number. This leads to the following theorem.
Let $G$ be a finite cyclic group of order $n$. Then the following classifications hold:
Let $G$ be a finite cyclic group of order $n$. By the definition of the cyclic subgroup sum, we have $S(G) = s(n)$, where $s(n)$ is the aliquot sum of $n$. The result then follows directly from the definitions of perfect, deficient, and abundant numbers. $ \qquad \blacksquare $
We've defined the cyclic subgroup sum of a finite cyclic group and explored its properties. We have shown that the cyclic subgroup sum can be used to classify finite cyclic groups based on whether their order is a perfect, deficient, or abundant number. This classification could potentially provide insight into the structure of finite cyclic groups and their subgroups. Here are several conjectures that could be explored in future work:
If $G$ is a finite cyclic group of order n and its cyclic subgroup sum $S(G) = n$ then for every proper subgroup $H$ of $G$, $S(H) \neq |H|$.
If $G$ is a finite cyclic group of order n and $H_1$ and $H_2$ are distinct proper subgroups of $G$, then $S(H_1) \neq S(H_2)$.
If $G$ is a finite cyclic group of order n and $H$ is a proper subgroup of $G$, then $G$ and $H$ do not share the same classification (deficient, perfect, or abundant) based on their cyclic subgroup sums.
If $G$ is a finite cyclic group with $S(G) = |G|$, then $Z_2$ is isomorphic to a subgroup of $G$.
There are infinitely many finite cyclic groups $G$ such that $S(G) = |G|$.
[1] Hardy, G. H., & Wright, E. M. (1979). An Introduction to the Theory of Numbers (5th ed.). Oxford University Press.
[2] Dummitt, D.S. and Foote, R.M. (2004) Abstract Algebra. 3rd Edition, John Wiley & Sons, Inc.
[3] Gallian, J. (2021). Contemporary Abstract Algebra (10th ed.). Chapman and Hall/CRC