TITLE: On the sum of the subgroup orders of a finite group
QUESTION [13 upvotes]: 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?

REPLY [4 votes]: Partial answer:

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.