Automorphisms of $\langle \mathbb{N}, \cdot \rangle$

Question

It is an elementary fact that multiplication in $\mathbb{N}$ is commutative:

$$(\forall n,m)\ n \cdot m = m \cdot n$$

This - among other things - implies that the representation of an $n \in \mathbb{N}$ as a product of primes (which always exists) is unique upto permutations of the factors in the term representing $n$.

But what does this have to do with the fact, that each permutation $\pi:\mathbb{N} \rightarrow \mathbb{N}$ that only permutes primes (as members of $\mathbb{N}$, not as members of a term!) induces an automorphism of $\langle \mathbb{N}, \cdot \rangle$ (as I learned here).

I assume that a permutation $\pi:\mathbb{N} \rightarrow \mathbb{N}$ induces an automorphism of $\langle \mathbb{N}, \cdot \rangle$ when

$$(\forall n,m)\ \pi(n)\cdot \pi(m) = \pi(n\cdot m)$$

What has to be shown then is:

When for a permutation $\pi:\mathbb{N} \rightarrow \mathbb{N}$

$$(\forall n)\ \pi(n) \neq n \rightarrow \operatorname{prim}(n)$$

then

$$(\forall n,m)\ \pi(n)\cdot \pi(m) = \pi(n\cdot m)$$

How do I show this even for the most simple permutation $\pi_{23} = (2,3)$? Is the above maybe the wrong definition of automorphism of $\langle \mathbb{N}, \cdot \rangle$? I am confused by the fact

$$\pi_{23}(2) \cdot \pi_{23}(2) = 3 \cdot 3 = 9 = \pi_{23}(9) \neq \pi_{23}(2 \cdot 2)$$

Answer 1

The $\pi_{23}$ you've written is only the permutation on the primes; it's not the induced permutation on $\mathbb{N}$. To get that one (call it $\Pi_{23}$) you need to use your definition to expand it to the non-primes. Then $\Pi_{23}(4)$ $= \Pi_{23}(2\cdot 2)$ $= \Pi_{23}(2)\cdot\Pi_{23}(2)$ $=\pi_{23}(2)\cdot\pi_{23}(2)$ $=3\cdot 3=9$ by definition. Essentially, if you write every $n$ as a vector 'over' the primes, with coefficients $\geq 0$ and all sufficiently large coefficients $0$, then the original permutation $\pi$ corresponds to a relabeling of the primes, and the induced permutation $\Pi$ corresponds to how numbers map under this relabeling. In this notation, starting with e.g. $120=2^4\cdot 3^1\cdot 5^1 = \langle 4,1,1\rangle$, we 'relabel' (reinterpret) the vector as $3^4\cdot 2^1\cdot 5^1 = 810$, so $\Pi_{23}(120)=810$.

Answer 2

(I denote the set of primes by $\mathbb{P}$.)

But what does this have to do with the fact, that each permutation $\pi:\mathbb{N} \rightarrow \mathbb{N}$ that only permutes primes (as members of $\mathbb{N}$, not as members of a term!) induces an automorphism of $\langle \mathbb{N}, \cdot \rangle$ (as I learned [here][1]).

[emphasis mine] The answer you linked to did not say that precisely. The key here is that a permutation of primes $\pi:\mathbb{P}\to\mathbb{P}$ induces an automorphism of $\mathbb{N}$. The induced automorphism is $\sigma$ for which $$ \sigma(p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}) = \pi(p_1)^{a_1} \pi(p_2)^{a_2} \cdots \pi(p_n)^{a_n}. $$

Your permutation $\pi_{23}$ is not induced by a permutation of primes in this way.