First question:

1) Is the sum of subgroup indices of dihedral group with $2n$ elements equal to $\sigma_2(n)+2\cdot \sigma(n)$?

Second question:

2) Is $\sigma_2(n)+2\cdot \sigma(n) \le L(H(D_n))$?

where $L(x) = x+\exp(x)\cdot \log(x)$, $H(D_n) = H_n-5/2+2 H_{n/2+1}+2/(n+4)$, if $n\equiv 0 \mod(2)$ and $H(D_n) = H_n-5/2+2 H_{(n+3)/2}$, if $n\equiv 1 \mod(2)$

The inequality might be seen as the Lagarias inequality for the dihedral group or if you want "the Riemann hypothesis for the dihedral group", where the "usual" Rieman hypothesis is for the cyclic group (Lagarias inequality is equivalent to RH)

Notation: $H_n=$ $n$-th harmonic number, $\sigma(n) = $ sum of divisors, $\sigma_2(n) = $ sum of squared divisors.

Context: Replace $\sigma(G)=\sum_{U \le G} |U|$ in the following question with $\sigma(G) = \sum_{U \le G} [G:U]$ and set $G=D_n = <r,s>, S= \{r,s\}$


For the second question, it seems numerically that the upper bound imposed on $\sigma(n)$ is bigger than the Lagarias upper bound, so the 2) question does not imply RH.

Edit: Here is some SAGE code which tests the two questions:

# Lee norm on Z/(n)
def lee(a,n):
    return min(a%n,(-a)%n)

# harmonic numbers:
def harmonic(n):
    return sum([1/k for k in range(1,n+1)])

# Harmonic numbers for DihedralGroup(n)
def HDn(n):
    return harmonic(n)+sum([1/(lee(a,n)+2) for a in range(n)])

# Harmonic numbers for DihedralGroup(n)
def HDn2(n):
    if n%2 ==0:
        h = 1/2+2*(harmonic(n/2+1)-3/2)+2/(n+4)
        h = 1/2+2*(harmonic((n+3)/2)-3/2)
    return harmonic(n)+h

# sum of subgroup indices:
def sigmaGr2(G):
    return sum([len(G)/len(U.list()) for U in (G.subgroups())])

# conjectured sum of subgroup indices for DihedralGroup(n):
def sigmaDihedral(n):
    return sigma(n,2)+2*sigma(n)

for n in range(1,21):
    h = HDn2(n)
    s = sigma(n,2)+2*sigma(n)
    S = sigmaGr2(DihedralGroup(n))
    print n, S == s  and s <= (h+exp(h)*log(h)).N()

Related question for cyclic group and $S=\{\pm 1\}$: https://mathoverflow.net/questions/331228/a-question-about-lagarias-inequality

Second Edit:

In Theorem 3.1 by Keith Conrad it is stated, that:

(a) For each $d|n$ there is a subgroup with index $2d$.

(b) For each $d|n$ there are $d^2$ subgroups with index $d$.

From this it follows that:

$$\sigma(D_n) = \sum_{d|n} 2d + \sum_{d|n} \sum_{i=0}^{d-1} d = 2 \sigma(n) + \sigma_2(n)$$ which was to be shown.

Question 2) remains open.

  • $\begingroup$ 1. should follow using the result and methods from this post. I obtain an equation in $n,\tau(n),\sigma(n)$ and still struggle with $\sigma_2(n)$. $\endgroup$ – Dietrich Burde May 13 at 9:25

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