# Zappa-Szép products of the group of integers with itself

Since my previous question didn't get much attention and I couldn't make any relevant progress on it, I thought it would be a good idea to "simplify" it by replacing monoids by groups. That is:

## Question

What are all the bijective functions $\alpha , \beta : \mathbb{Z} \to \mathbb{Z}$ satisfying the following functional equations?

$\bullet ~~~~ \alpha(0)=0, \quad \beta(0)=0\\ \bullet ~~~ \alpha^n(m+m')=\alpha^n(m) + \alpha^{\beta^m(n)}(m')\\ \bullet ~~~ \beta^m(n+n') = \beta^m(n) + \beta^{\alpha^n(m)}(n')$

Here, $\alpha^n$ denotes the $n$-fold composition of $\alpha$. So far I have only found the following solutions:

1) $\alpha(n)=\beta(n)=n$

2) $\alpha(n)=n$, $\beta(n)=- n$ (and like

2') $\alpha(n)=- n$, $\beta(n)=n$

3) $\alpha(n)=\beta(n)=-n$

I am not even sure if there are more solutions.

## Background

This is not necessary to understand the question, but I hope that it shows why the question is interesting.

The Zappa-Szép product (aka bicrossed product) $A \bowtie_{\sigma} B$ of two groups $A,B$, written additively for our purposes, with respect to a map $\sigma=(\sigma_1,\sigma_2) : U(B) \times U(A) \to U(A) \times U(B)$ (here $U(A)$ denotes the underlying set of $A$), is a group with underlying set $U(A) \times U(B)$, unit $(0,0)$, and multiplication $(a,b) +_{\sigma} (a',b') = (a + \sigma_1(b,a'),\sigma_2(b,a') + b')$. For the group axioms to be satisfied, the maps $\sigma_1,\sigma_2$ have to satisfy several requirements (see the links).

My goal is to classify all Zappa-Szép products of $(\mathbb{Z},+,0)$ with itself. Has this already been done in the literature? Since here an action is simply determined by a bijective map, this comes down to the system of functional equations as described above: We have $\alpha^n(m)=\sigma_1(n,m)$ and $\beta^m(n)=\sigma_2(n,m)$.

The groups corresponding to the solutions so far are the groups $\mathbb{Z} \times \mathbb{Z}$, $\mathbb{Z} \rtimes_{-1} \mathbb{Z} \cong \mathbb{Z} \ltimes_{1} \mathbb{Z} \cong \langle x,y : xyx^{-1}=y^{-1} \rangle$ (this is a well-known group) and $\langle x,y : yx=x^{-1} y^{-1},\, y x^{-1} = xy^{-1} \rangle$ (this group is mentioned in Cohn's article).

This paper remarks that the Zappa-Szép products of two finite cyclic groups have not been classified yet, although Jesse Douglas studied them in a couple of papers. They show that the Zappa-Szép products are simply semidirect products if one of the cyclic groups has prime order.

• I think my question is answered by Redei's paper link.springer.com/article/10.1007/BF02022554 . The classification was done under an assumption, which was proven by Cohn link.springer.com/article/10.1007/BF01899562 , which in turn used results by Ito eudml.org/doc/169505 . – HeinrichD Feb 23 '17 at 17:44
• Redei's paper is almost impossible to read: He uses expressions like ${\mathrm{A}}^{\mathsf{A}}$ where the elements ${\mathrm{A}},\mathsf{A}$ belong to different groups! And the reader is supposed to see (guess ...) to which group. – HeinrichD Feb 23 '17 at 18:31
• I haven't understood Redei's paper yet, but the only remaining solution seems to be $\alpha(n)=-n$ and $\beta(n)=n + 2 \cdot (n \bmod 2)$ (or with $\alpha,\beta$ exchanged), i.e. $\sigma(n.m) = ( m\cdot (-1)^n , n + 2 \cdot m \cdot (n \bmod 2) )$. The group has the presentation $\langle x,y : y x^{-1} = x y^{-1},\, y x = x^{-1} y^3 \rangle$. – HeinrichD Feb 23 '17 at 19:45