# On the sum of the subgroup orders of a finite group

Let $$G$$ be a finite group. Consider the function providing the sum of the subgroups orders $$\sigma(G) = \sum_{H \le G} |H|.$$ Note that if $$C_n$$ is cyclic of order $$n$$ then $$\sigma(C_n) = \sigma(n)$$, with $$\sigma$$ the usual divisor function. Consider the functions $$\sigma_{-}(n)= \min_{|G|=n} \sigma(G), \ \ \ \ \ \ \sigma_{+}(n)= \max_{|G|=n} \sigma(G).$$

This post is about a characterization of the extremizers, i.e. the finite groups $$G$$ such that $$\sigma(G) = \sigma_{\pm}(|G|)$$.

$$\begin{array}{c|c} n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline \hline \sigma_{-}(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24\newline \hline \sigma(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24 \newline \hline \sigma_{+}(n)&1&3&4&11&6&16&8&51&22&26&12&60&14&36&24 \end{array}$$

We can observe in above table that $$\sigma_{-}(n) = \sigma(n)$$, and it holds for all $$n < 256=2^8$$ (by checking on GAP).

Question 1: What are the finite groups $$G$$ such that $$\sigma(G) = \sigma_{-}(|G|)$$? Exactly the cyclic groups?

Next, consider the prime factorization of $$n$$ $$n=\prod_{i=1}^r p_i^{n_i},$$ then, candidates which come in mind for $$\sigma(G) = \sigma_{+}(|G|)$$ are the product of prime order cyclic groups: $$G = \prod_{i=1}^r C_{p_i}^{n_i}.$$ It works often but not always, as $$\sigma(S_3) = \sigma_{+}(|S_3|) = \sigma_{+}(6) = 16$$ whereas $$\sigma(C_2 \times C_3) = 12$$; but $$S_3 = C_3⋊C_2$$, moreover, for $$n \le 60$$, all the models I found are semi-direct product of prime order cyclic groups.

Question 2: What are the finite groups $$G$$ such that $$\sigma(G) = \sigma_{+}(|G|)$$?

Are there semi-direct product of prime order cyclic groups? Or at least supersolvable?

Here is a small remark: let $$n$$ be a number which is not abundant (that is, the sum of the proper divisors of $$n$$ is at most $$n$$). Suppose further that there is a non-cyclic finite group $$G$$ of order $$n$$. I claim that $$\sigma(G) > \sigma(n)$$.

Since $$G$$ is not cyclic, $$\langle x \rangle$$ is a proper subgroup of $$G$$ for each non-identity element $$x \in G$$. Hence each $$x \in G$$ is contained in some maximal (proper) subgroup of $$G.$$

Note also that $$G$$ has more than one maximal subgroup ( for if there is only one maximal subgroup $$M$$ of $$G$$, then as noted above, $$M$$ must contain each non-identity element of $$G$$, and it certainly contains the identity. This contradicts the fact that $$M$$ is proper).

Hence if we add the order of all maximal subgroups of $$G$$, we obtain an integer greater than $$|G|$$, since we count each non-identity element of $$G$$ at least once, and we count the identity element more than once. Hence we have $$\sigma(G) > 2n \geq \sigma(n)$$.

In particular, if $$n$$ is an integer greater than one which is not abundant, then we do have $$\sigma_{-}(n) = \sigma(n)$$, and only for cyclic groups $$G$$ of order $$n$$ do we attain $$\sigma(G) = \sigma(|G|).$$ This answers question 1 for groups $$G$$ whose orders are not abundant numbers. Furthermore, it shows that $$\sigma_{-}(n) = \sigma(n)$$ for any integer $$n >1$$ which is not abundant.

• Good! Your redaction can be shorten by writting: $G$ non-cyclic iff $G$ equals the union of its maximal subgroups (which share the trivial element). Then for $G$ non-cyclic, the sum of the order of its maximal subgroups must be greater than $|G|$, so that $\sigma(G) > 2|G|$ . – Sebastien Palcoux Jun 29 at 10:40
• Another way to get the same bound: every element is contained in the subgroup it generates and in the whole group (and these are different if $G$ is not cyclic). (By double counting, $\sigma(G)/|G|$ is the average number of groups containing a given element. I haven't been able to put this method to any further use.) – R. van Dobben de Bruyn Jul 1 at 0:14

The answer to Q1 is "yes" if one restricts $$G$$ to supersolvable groups or more generally groups that satisfy the converse of Lagrange's theorem.
If $$G$$ has the property that for every divisor of the group order there is at least one subgroup of that order, then in particular $$\sigma(G)\geq\sigma(|G|)$$. Furthermore, if $$\sigma(G)=\sigma(|G|)=\sigma(\mathbb{Z}/|G|)$$ holds, then there is exactly one subgroup of every order. In particular, all sylow groups are normal so that the group is the direct product of its sylows and inside every sylow subgroup, there is exactly one maximal subgroup so that it is cyclic.
Another observation: $$\sigma$$ is "multiplicative" for groups similar to how $$\sigma$$ is multiplicative for groups, i.e.
$$gcd(|G|,|H|)=1 \implies \sigma(G\times H)=\sigma(G)\cdot\sigma(H)$$
This follows from the fact that in these situations, every subgroup of $$G\times H$$ is uniquely decomposable as $$G_0\times H_0$$ with $$G_0\leq G$$ and $$H_0\leq H$$.
More generally, one can observe that $$\sigma(G\rtimes H) = \sum_{G_0\leq G} |G_0| \cdot\sigma(N_H(G_0))$$ holds for $$gcd(|G|,|H|)=1$$, because in this case every subgroup uniquely decomposes as $$G_0\rtimes H_0$$ by the Schur-Zassenhaus theorem. The right hand side is less than or equal to $$\sigma(G)\cdot\sigma(H)$$ with equality iff all subgroups of $$G$$ are $$H$$-invariant.