# Is the identity function a unique multiplicative homeomorphism of $\mathbb N$?

Endow the set $\mathbb N$ of positive integers with the topology $\tau$ generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $\mathbb N_0=\{0\}\cup\mathbb N$, where $a,b\in\mathbb N$. This topology is often referred to as the Furstenberg topology or the profinite topology. The space $\mathbb N_\tau:=(\mathbb N,\tau)$ is homeomorphic to the space $\mathbb Q$ of rational numbers (being a second-countable regular countable space without isolated points). So $\mathbb N_\tau$ has many non-trivial homeomorphisms. But I know no non-trivial homeomorphism of $\mathbb N_\tau$ which would be multiplicative is the sense that $f(x\cdot y)=f(x)\cdot f(y)$ for any $x,y\in\mathbb N$.

Problem 1. Is the identity function a unique multiplicative homeomorphism of $\mathbb N_\tau$?

An affirmative answer to this problem follows from an affirmative answer to

Problem 2. Are there prime numbers $a,b,c$ such that for any $d\in\mathbb N$ there exists $n\in\mathbb N$ such that $a^n\equiv 1\!\!\!\mod\! d\;\;$ but $\;\;b^n\not\equiv 1\!\!\!\mod\! c$?

• I'm always confused when someone calls $\mathbb{N}$ the set of positive integers... (+1 to the question, anyway!) – Qfwfq Sep 9 '18 at 10:50
• @Qfwfq Thank you for your +1. Concerning the notations, I arrived to the following compromise: $\omega=\{0,1,2,\dots\}$ is the set of finite ordinals but $\mathbb N$ of positive integers. Otherwise, it will be not faire: two notations for the set $\{0,1,2,3,\dots,\}$ and none for $\{1,2,3,\dots\}$. Moreover, according to the Eastern Ortodox traditions, we start counting from 1, not from 0. So, the lowest floor in our country is the 1-st floor, not the ground floor. So, for us (at least) the set $\{1,2,3,\dots\}$ for natural numbers is more natural than $\{0,1,2,3,\dots\}$. – Taras Banakh Sep 9 '18 at 13:21
• @Qfwfq The history of mathematics witnesses that for long time zero was not considered to be a (natural) number. It simply did not exist. – Taras Banakh Sep 9 '18 at 13:24
• I see your pont, ad it makes sense... It's just a habit and psychological think for me ;) – Qfwfq Sep 9 '18 at 15:38

No.

First observe that the automorphisms of the semigroup $\mathbf{N}^*$ (which you denote $\mathbb{N}$) are induced by permutations of primes.

Consider the automorphism $f$ induced by the transposition $(2,3)$ (thus, mapping $2^a.3^b.c$ to $2^b.3^a.c$, $c$ coprime to 6).

I claim that $f$ is continuous. Indeed, consider any convergent sequence $m_i\to m$. Thus, $m_i=m+r_i$ with $r_i$ tending to 0 in the profinite completion of $\mathbf{Z}$ (that is, for every $n\ge 1$ there exists $i_0$ such that $n$ divides $r_i$ for all large $i$).

Write $m=2^a.3^b.c$ with $c$ coprime to 6. There exists $i_0$ such that $2^{a+1}3^{b+1}$ divides $r_i$ for all $i\ge i_0$. So, for $i\ge i_0$, $m_i=2^a.3^b.c+2^{a+1}3^{b+1}.t_i$ for some $t_i$. So $m_i=2^a3^b(c+6t_i)$ with $c$ coprime to 6. Thus $f(m_i)=2^b3^a(c+6t_i)=f(m)+r'_i$, with $r'_i=2^{b-a}3^{a-b}r_i$. Thus $r'_i$ also tends to 0 in the profinite completion of $\mathbf{Z}$. Thus $f(m_i)$ tends to $f(m)$. Hence, $f$ is continuous.

The argument adapts immediately to any finitely supported permutation of the set of primes.

• Remark: there exists a permutation of primes that induces a non-continuous monoid automorphism. Indeed, from Dirichlet's theorem, there exist two sequences of primes $(p_n)$, $(q_n)$ such that $p_n$ tends to 1 in the profinite completion, and $q_n$ equals 2 modulo 3. We can suppose they are injective with disjoint image. Let $t$ exchange $p_n$ and $q_n$ for all $n$, and induce $f$. Then $p_n\to 1$ while $f(p_n)=q_n$ does not tend to $1=f(1)$. – YCor Sep 8 '18 at 23:46
• Additional remark: there also exist infinitely supported permutations $\sigma$ (actually $2^{\aleph_0}$ many) of primes inducing a continuous monoid automorphism $f_\sigma$. Namely, consider a sequence of primes $(p_n)_{n\ge 0}$ tending to 1 in the profinite topology (as in my previous comment), and any permutation $\sigma$ supported on $\{p_n:n\ge 0\}$. Then $f_\sigma$ is continuous. This is not hard to check (one checks that in general, $f_\sigma$ continuous at 1 implies $f_\sigma$ continuous, and for such a choice of $\sigma$, one checks continuity at 1). I can give details upon request. – YCor Sep 9 '18 at 8:22
• But $\mathbb{N}^*=\beta\mathbb{N}\setminus\mathbb{N}$. – Ramiro de la Vega Sep 9 '18 at 12:41
• @RamirodelaVega I explicitly used $\mathbf{N}^*$ to mean $\{n\in\mathbf{Z}:n>0\}$; I'm aware that notation varies. – YCor Sep 9 '18 at 13:19
• It is interesting to remark that such "finitely supported" multiplicative biejctions of $\mathbb N$ are discontinuous in the Golomb topology (generated by the arithmetic sequences $a+b\mathbb N_0$) because of the density of primes in this topology. – Taras Banakh Sep 9 '18 at 13:34