Let $G$ be a finite group $S\subset G$ a generating set $|g|=$ word length with respect to $S$. Set

$$ \sigma(G) = \sum_{H \le G} [G:H]$$

Let $\rho$ be the regular representation and set $A_G := \sum_{g \in G} \frac{1}{1+|g|} \rho(g)$. Then $|A_G| = H_G = \sum_{g \in G} \frac{1}{1+|g|}$, $A_G$ is a normal matrix (and possibly not singular), and $A_G/H_G$ is a doubly stochastic matrix. Let $L(x) = x+\exp(x)\log(x)$ be the Lagarias operator. The "group-theoretic" Lagarias inequality might then be stated as:

$$\sigma(G) \le L(H_G) = L(|A|) (=^? |L(A)|)$$

where $|B| = $ spectral norm. (For $G=C_n$ the cyclic group and $S = \{a\}$ this is equivalent as shown by Lagarias to Riemann Hypothesis.) I have put an $?$ on the equality, since numerics suggest this is true, but I have no proof yet. For a possibly infinte group $G$, Marcus du Sautoy and Fritz Grunewald define:

$$\zeta_G(s) = \sum_{n=1}^\infty \frac{a(G,n)}{n^s} $$ where $a(G,n) = |\{H \le G : [G:H] = n\}|$. For $G = \mathbb{Z}$ this is the Riemann Zeta function. For $G= C_n$ the cyclic group we get: $$\zeta_{C_n}(s) = \sum_{d|n} \frac{1}{d^s} = \sigma_{-s}(n)$$ which was studied by Ramanujan to give (under the RH) an upper bound for $\sigma(n)$.

Now, what does that have to do with primes: Intuitively if
$$\hat{\pi}(n):= \{ p | p \text{ prime }, p \le n \}$$ then $\hat{\pi}($|G|$)$ will determine how much $\sigma(G)$ can grow at most, since by Lagranges theorem for each $H \le G$ we have $[G:H]||G|$, hence $[G:H]$ must be divisible by some primes $p \in \hat{\pi}(|G|)$. For instance by Sylows theorems, we get a tight ($G=C_{p^k}=$ cyclic p-group) lower bound for $\sigma(G)$:

$$\sigma(G) \ge |G| + |G| \sum_{1\le i \le r} \sum_{1 \le n_i \le \alpha_i} \frac{1}{p_i^{n_i}}$$

where $|G| = \prod_{1 \le i \le r} p_i^{\alpha_i}$ is the prime factorization.

So there seems to be a conjectured relationship between the Riemann Hypothesis, primes and finite groups.

My question is, what is the relationship between:

$$\zeta_{\mathbb{Z}}(s), \zeta_G(s) \text{ and } \zeta_{\mathbb{Z}\times G}(s)$$

where $G$ is a finite group and the group $\mathbb{Z}\times G := \mathbb{Z}\times_S G$ is defined in my previous question and is not the direct product. (It turns out, it is the direct product!)

Thanks for your help!

Edit 25.05.2019: All these upper bounds have something in common:

Let $G = C_n$ be the cyclic group.

1) Lagarias inequality: zeta_(Cn) ( -1) = sigma(n) <= L(H_n)

2) Ramanujans upper bound under RH for zeta_(Cn)(s) = sigma_{-s}(n)

3) There is also an upper bound equivalent to RH for the number of divisors:

zeta_(Cn)(0) = tau(n) <= .... some upper bound

4) The "group theoretic" conjectured Lagarias upper bound, can be interpreted as: zeta_G(-1) = sigma(G) <= L(H_G) . 

These are all upper bounds for the zeta functions of some finite group and have directly or indirectly something to do with RH.

Edit 04.06.2019: I found some (unpublished) upper bounds to $\zeta_{C_n}(s) = \sigma_{-s}(n)$ which are equivalent to Lagarias inequality and hence to RH.

  • 1
    $\begingroup$ The definition you give of $\zeta_G(s)$ is incorrect. Either it is $\sum_{H\subset G} 1/[G:H]^s$ with the sum running over $H$ of finite index or it is $\sum_{n\geq 1} a(G,n)/n^s$ where $a(G,n)$ is the number of subgroups of $G$ of index $n$. You have mixed together these different formulas by using $a(G,n)$ as the numerator of $1/[G:H]^s$ when $[G:H] = n$. Proofread your post. $\endgroup$
    – user1728
    May 23, 2019 at 9:38
  • $\begingroup$ yes. thanks. you are right. i will correct that in a moment $\endgroup$
    – user6671
    May 23, 2019 at 9:45
  • $\begingroup$ You start the question by mentioning a notation for a generating set $S$, the word length $|g|$, and an operator $A_G$, but these have nothing directly to do with posing the question you are asking. Consider removing that to make the content of the post less cluttered. $\endgroup$
    – user1728
    May 23, 2019 at 9:49
  • $\begingroup$ @user1728: Thank your for your comment. The word length $|g|$ has something to do with the group $\mathbb{Z}\times G$ and the matrix $A_G$ has something to do with Lagarias inequality ( = RH if $G$ is cyclic group). $\endgroup$
    – user6671
    May 23, 2019 at 10:05
  • $\begingroup$ Oh, $\mathbb Z \times G$ is not the direct product! Make that more explicit in the post. You have a link to your earlier question about that group but I still think you should emphasize this more clearly. $\endgroup$
    – user1728
    May 23, 2019 at 10:15

1 Answer 1


Unless I'm mistaken, the group that you're constructing here is an extension of $\mathbb{Z}$ by $G$. It would cause less confusion if you reserved the notation $\mathbb{Z} \times G$ for the direct product. A fairly common notation for this extension is $\mathbb{Z} : G$.

As far as your question on the relationship between zeta functions, a relationship of this type is worked out for any normal subgroup $N$ of a finite group $K$ in a paper of Ken Brown:

Brown, Kenneth S., The coset poset and probabilistic zeta function of a finite group, J. Algebra 225, No. 2, 989-1012 (2000). ZBL0973.20016.

Note that Brown's $P(G,s)$ is (at least in the case where $G = \mathbb{Z}$, haven't completely carefully checked elsewhere) the reciprocal of the $\zeta$ function. Brown also uses a theorem of Gaschütz which is stated only for finite groups. It seems likely to me, however, that you might be able to extend some of what Brown does to your situation.

  • $\begingroup$ Thanks for your answer. I will have a look at the paper! $\endgroup$
    – user6671
    May 23, 2019 at 14:22
  • $\begingroup$ I am reading the paper you suggested. Do you know why Kenneth S. Brown, does exclude $ H = G$ from the coset poset definition? $\endgroup$
    – user6671
    May 23, 2019 at 15:41
  • 4
    $\begingroup$ Because he's looking at the simplicial complex consisting of chains in the poset, and if you have a top element, it forms a cone (so is contractible). If you add on G and also \emptyset at the bottom, you indeed get a lattice. $\endgroup$ May 23, 2019 at 16:11
  • $\begingroup$ thanks for the clarification. i find the coset poset very interesting since $\sigma(G) = |G|+|\text{ coset posets of} G|$ $\endgroup$
    – user6671
    May 23, 2019 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.