What does the monoid quotient $\Bbb N^+/\langle3,4\rangle$ look like, where $\Bbb N^+$ is the multiplicative monoid produced by multiplying primes? In particular, is it by any chance relatively quick and easy to show that:
$\lvert 3x+2^{\nu_2(x)}\rvert\leq\lvert x\rvert$
Where $\lvert\cdot\rvert$ indicates the order in the quotient monoid? and $p^{\nu_p(x)}$ indicates the highest power of $p$ that factors $x$.
This is of course a quick and simple question to ask, but the motivation is somewhat involved. I believe it requires a measure of infinite order of a group, which I am unsure how to define but I think there is enough info below to do so. The motivation for the question is as follows - Collatz conjecture related:
If the Collatz conjecture is true then I think I can redefine the Collatz function as a closure operator on a totally disconnected subspace of $\Bbb R$. Each positive integer has infinite order, but its order is an ordinal number, which the Collatz function reduces.
If so, there is a torsion quotient monoid of $\Bbb N^+$ such that $\lvert 3x+2^{\nu_2(x)}\rvert\leq\lvert x\rvert$. Its identity is $3^n\cdot\{\{2^{2m}:m\in\Bbb Z\},\{2^{2m-1\}:m\in\Bbb Z\}\}$, and this is the motivation for the question.
For anyone interested in the detail (and in case it helps answer the question above), I'll elaborate on the definitions:
Working with the set $X=\Bbb Z[\frac16]>0$ with topology inherited from the real line, let the quotient $X/\langle2,3\rangle$ have unique representatives drawn from the 5-rough numbers.
Let $f(x)=x+\frac{21}{64}\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)}$
Then $f^n(x)=x+\left(1-2^{-6n}\right)\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$
Then we have $\overline F=\lim_{n\to\infty}f^n(x)=x+2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$
It is fairly easy to show that every $f^n$ is well-founded - the "zero" of every sequence is $2^{\nu_2(x)}\cdot3^{\nu_3(x)}\cdot p:p\equiv\{1,7,11,17,19,23\}\pmod24$
What does this have to do with the Collatz graph? Well, modulo (multiplicatively) the powers of $2$ and $3$, $f^n(x)$ is one of two open sets of the numbers sharing the same successor in the Collatz graph. Then $\overline F(x)$ is that successor. The reason for breaking into two open sets is that it improves behaviour of powers of $3$ in the quotient.
Then the Collatz conjecture claims that under transfinite composition $\lim_{n\to\omega^\omega} f^n(x)\to\frac{2^\omega}{3^\omega}$ or alternatively that for all $x$ there is some $p\in\Bbb N$ such that $\overline F_n(x)\in\langle2,3\rangle$
If the conjecture is true, each number converges to either the odd powers of $2$ or the even ones, divided by increasing powers of $3$. Hence it makes sense to think of $\langle3,4\rangle$ as the identity of the monoid and ask whether $\overline F(x)$ is akin to $x^\omega$ in a monoid quotient.
The multiplicative monoid $\Bbb{N}^+$ is the free commutative monoid over a countable set of generators, in this case, the set of primes $\Bbb{P}$.
According to your definition of the quotient by $\langle 3, 4 \rangle$, you get the quotient of the (multiplicative) free commutative monoid over the set $\Bbb{P} - \{3\}$ by the relation $2^2 = 1$. Thus you get a direct product of a free commutative monoid over a countable set of generators by the two-element cyclic group $C_2$ (since $2^2 \sim 1$).
I am not sure to understand the meaning of your inequation. In the quotient, every $n$ having a prime factor different from 2 and 3 will have infinite order, so your inequality is trivially satisfied (but does not make much sense). If $x = 3$, then the order of $x$ in the quotient is $0$, $2^{v_2(3)} = 1$, $3x + 2^{v_2(3)} = 10$ and $10$ has infinite order in the quotient.