What are the elements of $\Bbb Z[\frac16]/{\sim}$ and what do the subgroups and orders of elements look like?
In which $\exists i\in \Bbb Z:4^ia=b\implies a\sim b$
What's the identity element, for example?
The background is that I have reason to become interested in this as part of my study of the Collatz conjecture and I want to understand the elements better. I'm working towards the question:
Is there some obvious reason why in this group $\lvert3x+2^{\nu_2(x)}\rvert\leq\lvert x\rvert$?
As for my own attempt at an answer. Intuition tells me tentatively it may be the trivial group or possibly the cyclic group $2Z$, or may have $2Z$ as a subgroup, but I have no formal knowledge of how to show that so I'm keen to learn how group theory can be applied to this.
We have that $1\sim\frac14$ which gives us $2\sim5$ and probably gives us $2\sim5\sim42\sim85\sim682\cdots$ which (modulo the powers of $4$) is one half of the equivalence class of integers the same distance from $1$ by the function $f(x)=(3x+2^{\nu_2(x)})\lvert3x+2^{\nu_2(x)}\rvert_2$ - the other half being $1\sim10\sim21\sim170\sim341\cdots$
We have that $1\sim4$ also gives us that $5-2=4+1-2=0$ and therefore $3=0$ which again corresponds with the Collatz graph since $3x$ are the leaves the graph and can be ignored.
I'm particularly unclear however on how these relations will affect e.g. $\frac13$, and whether we can achieve $\forall x:x\sim2x$. I suspect $x\sim2x$ can't be achieved and the powers of $2$ remain partitioned into two alternating parts as per the above example.
Hint 1: A background in basic algebra will show that what you are factoring out by "making an equivalence relation such that $4x \sim x$ for all $x$ and such that factoring it out preserves the ring operations" is nothing else than the ideal $\langle (4-1) \rangle = \langle 3\rangle$.
Hint/Exercise 2, for a generalised solution:
Let $S$ be a set of primes, and $\Bbb Z [S^{-1}]$ the localisation of $\Bbb Z$ at $S$ ($\simeq$ the smallest subring of $\Bbb Q$ containing the inverses of all elements of $S$).
For $k \in \Bbb Z$ with prime factorisation $k = \pm \prod p_i^{n_i}$, let $k_S$ be the number $k$ with the prime factors which are in $S$ "divided out" (formally, set $k_S := \displaystyle \prod_{p_i \notin S} p_i^{n_i}$). Then: $$\Bbb Z [S^{-1}] /\langle k \rangle \simeq \Bbb Z/\langle k_S\rangle. $$
In the original question, $S = \lbrace 2,3\rbrace$ and $k=3$ leading to the trivial ring $\Bbb Z/\langle 1\rangle$. In the latest comment, $k=p-1$ and all you know about $S$ is that it contains $p$, which is not enough to give a definite answer.