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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.

Robert Frost
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    "which gives us $2 \sim 5$" -- why? There is no integral $i$ such that $2 \times 4^i = 5$. Are you suggesting there is algebraic structure on the quotient $\mathbb Z[\tfrac16]/\sim$? Why? – Mees de Vries Oct 02 '18 at 12:48
  • I'm not sure if this is what you want, but $\mathbb{Z}[1/6] \simeq\mathbb{Z}$ by multiplying by 6. Then the subgroup you are looking at in $\mathbb{Z}$ is $\langle 24 \rangle$; thus, your group is just cyclic with 24 elements. – JJC94 Oct 02 '18 at 12:50
  • @MeesdeVries My logic is that $1/4\sim1$. Then from $4\times 1/4+1=2$ and $4\times 1 + 1=5$ it follows that $2\sim5$. Is that logic flawed? – Robert Frost Oct 02 '18 at 13:59
  • It assumes that if $p$ is a polynomial and $x \sim y$ then also $p(x) \sim p(y)$, and in fact that is not true, since $2/5$ is not an integral power of $4$. – Mees de Vries Oct 02 '18 at 14:01
  • @MeesdeVries Yes, I see my error now. The quotient as I originally wrote it was contradictory and $\sim$ didn't define an equivalence class. I've adjusted it. – Robert Frost Oct 02 '18 at 14:10
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    Are you suggesting that $\sim$ is the least equivalence relation such that $a, b$ with $a/b = 4^i$ are related, and furthermore the algebraic operations are preserved? – Mees de Vries Oct 02 '18 at 14:13
  • @MeesdeVries yes, that may be a better wording. I guess I have that $\forall x:x+4x\sim x+x\implies 2x=5x$ – Robert Frost Oct 02 '18 at 14:21
  • @JJC94 do you mean my multiplying by $6$ up to infinitely many times? As e.g. we have $\frac1{18}\in\Bbb Z[\frac16]$ – Robert Frost Oct 02 '18 at 14:24
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    @RobertFrost then your ring is trivial, for you have $1/4 \sim 1 \implies 0 \sim 3/4 \implies 8/6 \times 0 \sim 8/6 \times 3/4 \implies 0 \sim 1$. – Mees de Vries Oct 02 '18 at 14:27
  • @MeesdeVries so my intuition was correct. Thank-you so much for your help. I may ask this as another question if it's not a trivial thing, but is there some "measure" of the "length" of the equivalence relation required to set $a\sim b$ in these circumstances? Or is that degenerate / convoluted? – Robert Frost Oct 02 '18 at 14:34
  • @RobertFrost, that would just be the length of a path from $a$ to $b$ in a graph, where the edges are given by the conditions you enforce. (The equivalence classes are then your connected components.) – Mees de Vries Oct 02 '18 at 14:40
  • Is $\mathbb{Z}[a]$ not the free abelian group generated by $a$? – JJC94 Oct 02 '18 at 23:50
  • @JJC94 it's the smallest subfield of $\Bbb Q$ that contains $\frac16$. Sorry if that wasn't clear. I'm copying conventions I've seen others use, but still learning when they might be ambiguous. – Robert Frost Oct 03 '18 at 03:43
  • @MeesdeVries I guess this generalises to every subfield of $\Bbb Q$ that contains $1/p^n$, then if $\forall x: x\sim px$ it is trivial? – Robert Frost Oct 03 '18 at 04:00

1 Answers1

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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.

Torsten Schoeneberg
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