Let $G$ be a finite group, $S \subset G$ a generating set. Set $\sigma(G):=\sum_{U \subset G} |U| $, where the sum runs over all subgroups $U$ of $G$. Set $H_G := \sum_{g \in G} \frac{1}{|g|+1}$, where $|g|:= $ word length (with respect to $S$). For $G:=\mathbb{Z}/(n)$ we get $\sigma(G) = \sigma(n)=$ sum of divisors of $n$. and $H_{\mathbb{Z}/(n)} = H_n=n$-th harmonic number, where $S=\{+1\}$. My naive conjecture inspired by Lagarias inequality is $$ \sigma(G) \le H_G + \exp(H_G) \log(H_G)$$

For $G:=\mathbb{Z}/(n)$ and $S:=\{+1\}$ this is the Lagarias inequality. I have checked in Sagemath for the symmetric group up to $n=6$:

```
def sigmaGr(G):
return sum([len(U.list()) for U in (G.subgroups())])
def wordLen(g):
return g.length()
def HG(G):
return sum([1/(wordLen(g)+1) for g in G.list()])
def LG(G):
H = HG(G)
return (H+exp(H)*log(H)).N()
for n in range(1,6):
G = SymmetricGroup(n)
print sigmaGr(G),LG(G)
```

My question is, if this inequality can be proved for the generating set $S:=G$ or if there are finite groups and generating sets for which this inequality is false?

For $S=G$ it is $H_G=(|G|+1)/2$ and for $G$ the cyclic group, the inequality reduces to

$$\sigma(n)\le (n+1)/2+\exp((n+1)/2)\log((n+1)/2)$$

so the question is if one can prove this inequality?

**Edit 24.05.2019**:
It seems that it is better to define $\sigma(G)$ as $= \sum_{H \le G} [G:H]$ which in the case of cyclic groups is equal to the first definiton $=\sigma(n)$. Also this notion of $\sigma(G)$ is related to the zeta function of the finite group $G$ as we have:

$$\zeta_G(-1) = \sigma(G)$$

where $$\zeta_G(s) = \sum_{H \le G} \frac{1}{[G:H]^s}$$