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Is $(\Bbb Z[\frac{1}{2}]/\Bbb Z)/{\sim}$ a torsion group?

let $\sim$ be the transitive completion of the equivalence relation $x\sim x+2^{v_2(x/4)}\text{ in }\Bbb Z[\frac{1}{2}]/\Bbb Z$

I think it is but don't really know where to start proving it. Maybe it's just a trivial fact.

I tried using a morphism to $\Bbb C$:

$$f(2^nx)=2^ne^{2xi\pi}$$

and to look at the behaviour of this to show that this stabilises to $2^ne^0$ under multiplication.

But I'm inexperienced in group theory. It seems likely to me the result may just be a trivial fact to somebody more experienced.

Just to more clearly define the equivalence relation I will give the closed form. It was tricky to write this but I think there is no error:

$x\sim y\iff \quad\exists n\in\Bbb Z:y=x+\dfrac{4^n-1}{3\cdot4^n\lvert x\rvert_2}$

Robert Frost
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    I changed $$(\Bbb Z[\tfrac{1}{2}]/\Bbb Z)/\sim,$$ coded as (\Bbb Z[\frac{1}{2}]/\Bbb Z)/\sim, to $$(\Bbb Z[\tfrac{1}{2}]/\Bbb Z)/{\sim},$$ coded as (\Bbb Z[\frac{1}{2}]/\Bbb Z)/{\sim}. The former has too much space to the left of $\text{“}\sim\text{''}$ because that symbol is usually a binary relation symbol, and so has that amount of space before it. But in this case it refers to an object. – Michael Hardy Jun 25 '18 at 17:39
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  • "transitive completion" of an equivalence relation means nothing. You possibly mean transitive completion of an incidence relation. Second, for the quotient to be a group you need to mod out by a subgroup. Maybe you kill the $2^{v_2(x/4)}$... anyway you start from a torsion group, so asking whether one of its quotients (assuming it makes sense) is torsion sounds like a tautology.
    • – YCor Jun 25 '18 at 19:01
  • @YCor If I understand correctly you're saying that the relation being a quotient means I'm right thinking "Maybe it's just a trivial fact." (that it's a torsion group). – Robert Frost Jun 25 '18 at 20:21
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    Better don't rely on me and prove it as an exercise. – YCor Jun 25 '18 at 20:24
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    You don't need go that far. You have to prove two independent easy things: (a) $Z[1/p]/Z$ is a torsion group (b) every quotient group of a torsion group is a torsion groups. – YCor Jun 25 '18 at 20:49
  • Thank-you @YCor for your help. I already know $\Bbb Z[\frac{1}{2}]/\Bbb Z$ to be torsion. I'm not experienced enough in group theory to know every equivalence class creates a quotient. – Robert Frost Jun 25 '18 at 21:29
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    It's the very basics of group theory, the notion of quotient group. – YCor Jun 25 '18 at 22:00
  • @YCor there're disputing your claim here: https://math.stackexchange.com/q/2833459/334732 . What they're saying looks legit. Do you have any comment? (it's be better more me if your claim was true!!) – Robert Frost Jun 27 '18 at 12:03
  • @Ycor just to be clear, more than a) and b) are required. It's also necessary to show c) that the resulting quotient continues to obey the group axioms. – Robert Frost Oct 09 '18 at 06:40