# A zeta function for the Klein Four group?

Let $$G$$ be a finite group, $$S \subset G$$ a generating set, $$|g|:=|g|_S=$$ word-length with respect to $$S$$. Let $$\phi(g,h)=|g|+|h|-|gh| \ge 0$$ be the "defect-function" of $$S$$. The set $$\mathbb{Z}\times G$$ builds a group for the following operation:

$$(a,g) \oplus (b,h) = (a+b+\phi(g,h),gh)$$

On $$\mathbb{N}\times G$$ is the "norm": $$|(a,g)| := |a|+|g|$$ additive, which means that $$|a \oplus b| = |a|+|b|$$. Define the multiplication with $$n \in \mathbb{N_0}$$ to be:

$$n \cdot a := a \oplus a \oplus \cdots \oplus a$$

(if $$n=0$$ then $$n \cdot a := (0,1) \in \mathbb{Z} \times G$$).

A word $$w := w_{n-1} w_{n-2} \cdots w_0$$ is mapped to an element of $$\mathbb{Z} \times G$$ as follows:

$$\zeta(w) := \oplus_{i=0}^{n-1} (m^i \cdot (0,w_i))$$

where $$m := \min_{g,h\in G, \phi(g,h) \neq 0} \phi(g,h)$$.

We let $$|w|:=|\zeta(w)|$$ and $$w_1 \oplus w_2:=\zeta(w_1)\oplus \zeta(w_2)$$

Then we have $$|w_1 \oplus w_2| = |w_1|+|w_2|$$.

For instance for the Klein four group $$\{0,a,b,c=a+b\}$$ generated by $$S:=\{a,b\}$$, we get sorting the words $$w$$ by their word-length:

$$0,a,b,c,a0,aa,ab,ac,b0,ba,bb,bc,c0,ca,cb,cc,a00,a0a,a0b,a0c$$

corresponding to the following $$\mathbb{Z}\times K_4$$ elements $$\zeta(w)$$:

$$(0,0),(0,a),(0,b),(0,c),(2,0),(2,a),(2,b),(2,c),(2,0),(2,a),(2,b),(2,c),(4,0),(4,a),(4,b),(4,c),(4,0),(4,a),(4,b),(4,c)$$

corresponding to the the following "norms" of words $$|w| = |\zeta(w)|$$:

$$0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6$$

Let $$a_n, n\ge 0$$ be the sequence of numbers generated by the Klein four group.

1) Is $$\sum_{n=1}^\infty \frac{1}{a_n^s} = \sum_{n=1}^\infty \frac{n+1}{n^s} = \zeta(s-1) + \zeta(s)$$ where $$\zeta$$ denotes the Riemann zeta function?

I have checked this with SAGE math up to a certain degree and it seems plausible, however I have no idea how to prove it.

2) Is every $$a_n$$ the product of primes $$p=a_k$$ for some $$k\le n$$?

3) Let $$\pi_{K_4}(n) = |\{ k : \text{a_k is prime, k \le n}\}|$$ be the prime counting function of the sequence. What is the approximate relationship to the usual prime counting function $$\pi(n)$$?

• What is your definition of $a_n$? – Thomas Browning Jan 28 at 3:39
• @ThomasBrowning It is the sequence for the Klein group defined in the previous question. – user6671 Jan 28 at 3:49
• I don't see an integer sequence defined in the previous question – Thomas Browning Jan 28 at 3:53
• @ThomasBrowning mathoverflow.net/a/351113/6671 sorry it was defned in the answer – user6671 Jan 28 at 3:54
• Are you talking about this sequence: 0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6? This sequence seems to depend upon a lexicographic ordering of words. Is this correct? – Thomas Browning Jan 28 at 4:02

Consider a word $$w=w_0w_1\cdots w_{n-1}$$ with each $$w_i\in\{0,a,b,c\}$$. Following your notation, $$\lvert w\rvert=\lvert\zeta(w)\rvert=\left\lvert\bigoplus_{i=0}^\infty m^i\cdot(0,w_i)\right\rvert=\sum_{i=0}^{n-1}\lvert 2^i\cdot(0,w_i)\rvert=\sum_{i=0}^{n-1} 2^i\lvert w_i\rvert$$ where $$\lvert0\rvert=0$$, $$\lvert a\rvert=1$$, $$\lvert b\rvert=1$$, $$\lvert c\rvert=2$$. We now consider the generating function $$F_n(x)=\sum_wx^{\lvert w\rvert}$$ where the sum is over all words $$w=w_0w_1\cdots w_{n-1}$$. Then \begin{align*} F_n(x)&=\sum_{w_0}\sum_{w_1}\cdots\sum_{w_{n-1}}x^{|w_0|+2|w_1|+\cdots+2^{n-1}|w_{n-1}|}\\ &=\left(\sum_{w_0}x^{|w_0|}\right)\left(\sum_{w_1}x^{2|w_1|}\right)\cdots\left(\sum_{w_{n-1}}x^{2^{n-1}|w_{n-1}|}\right)\\ &=(1+2x+x^2)(1+2x^2+x^4)(1+2x^4+x^8)\cdots(1+2x^{2^{n-1}}+x^{2^n})\\ &=(1+x)^2(1+x^2)^2(1+x^4)^2\cdots(1+x^{2^{n-1}})^2\\ &=(1+x+x^2+\cdots+x^{2^n-1})^2\\ &=1+2x+\cdots+(2^n-1)x^{2^n-2}+2^nx^{2^n-1}+(2^n-1)x^{2^n}+\cdots+x^{2^{n+1}-2}. \end{align*}
What this shows is that among the first $$4^n$$ terms of the sequence: \begin{align*} &0\text{ appears }1\text{ times},\\ &1\text{ appears }2\text{ times},\\ &2\text{ appears }3\text{ times},\\ &\cdots\\ &2^n-2\text{ appears }2^n-1\text{ times},\\ &2^n-1\text{ appears }2^n\text{ times},\\ &2^n\text{ appears }2^n-1\text{ times},\\ &\cdots\\ &2^{n+1}-4\text{ appears }3\text{ times},\\ &2^{n+1}-3\text{ appears }2\text{ times},\\ &2^{n+1}-2\text{ appears }1\text{ times}.\\ \end{align*} This resolves question 1. Since this seems to be purely a combinatorics question, I would not expect the second and third questions to have particularly interesting answers.
I will remark that $$\pi_{K_4}(4^n)=\sum_{p\leq2^{n+1}-2}(2^n-\lvert p-2^n+1\rvert)p$$ so you could use some analytic number theory to get asymptotics for $$\pi_{K_4}(n)$$ if you wanted.