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14:06
I may just have the transitivity relation you seek if you were available briefly for a chat in relation to this question: mathoverflow.net/questions/209373 . Mine hangs around the number $21$ rather than $42$ but it holds for $2^i\cdot21:i\in\Bbb Z$ (and $3^j\cdot$ too for that matter). But I'm not versed enough in your terminology to work out if what we each have are complementary components to each other's work, unrelated or equivalent.
 
3 hours later…
17:17
@RobertFrost Sorry, I don't understand. -- Can you perhaps explain what you mean?
@StefanKohl You ask if the group action is transitive on $\Bbb N_0$
Do you have CHatjax installed?
I have a relation on the Collatz graph which you may be able to use to show the transitivity you need to prove the Collatz conjecture, or it may not be any use to you.
I think you will very quickly be able to tell.
In your question which I linked, you ask this: Question (new version): Let G<Sym(Z) be a group generated by 3 class transpositions, and let m be the least common multiple of the moduli of the residue classes interchanged by the generators of G. Assume that G does not setwisely stabilize any union of residue classes modulo m except for ∅ and Z, and assume that the integers 0,…,42 all lie in the same orbit under the action of G on Z. Is the action of G on N0 necessarily transitive?
Then I ask whether the following relation is any use to you in proving this.
@RobertFrost Which relation do you mean? (Besides, I don't have Chatjax installed).
@StefanKohl Ok this is the relation. I'll do my best without Chatjax:
Are you aware the 5-rough positive integers are a sufficient set to prove the Collatz conjecture (it affects how I write the relation)
Ok I'll write it using N_0 rather than the 5-rough integers (which would have been marginally simpler)
@RobertFrost Basically any residue class suffices (but as far as I can tell, this won't help).
Connectedness in a graph is an equivalence relation so I will write \sim to indicate connectedness
Then obviously we have the relation that $x\sim3x+2^{\nu_2(x)}$ and basically the Collatz conjecture states that every integer is $\sim$ some sufficiently high power of 2
The relation which I tohught might be of use to you is this:
$x\sim 21(3x+2^{\nu_2(x)})$
If we look at the two smallest immediate predecessors of any odd number, i.e. 1,5 are the immediate predecessors of 1
Then the relation x+21(3x+1) generates all the other immediate odd predecessors from these two.
Sorry that: chat.stackexchange.com/transcript/message/47710015#47710015 is wrong, it should say $x\sim x+21(3x+2^{\nu_2(x)})$
17:33
Do you denote by \nu_p(x) the p-adic valuation of x?
yes, but you can make this a lot easier to think about by thinking of that as just the number 1 if you restricted yourself to the odd numbers (which you can do since the 5-rough numbers are a sufficient set to prove the Collatz conjecture)
It's also better to forget about the multiples of 3 as these are the leaves of the graph.
What I was particularly curios about is that for you the number 42 is important, and for me the number 21\cdot2^n:n\in\Bbb Z is important.
Pls let me know if anything I say makes any sense and if not, where to disambiguate. Sadly I have a habit of being difficult to understand
@RobertFrost I think what concerns 42 vs. 21, we are really talking about two different things (the 42 in my question as such has no relation whatsoever to the Collatz conjecture -- it is only relevant for the generalization).
@StefanKohl Unless of course the maths behind the Collatz conjecture has a role in the generalisation.
17:51
@RobertFrost It quite likely has -- but not on that level ... . Besides -- do you see any reason why the relation $x\sim x+21(3x+2^{\nu_2(x)})$ might help more than just any such relation read off from the tree of predecessors?
@StefanKohl oh yes, this is an important relation
It has a few properties:
Firstly a touch of housekeeping
Define the equivalence relation $x\sim y\iff \exists i\in\Bbb Z:2^i\cdot x=y$
i.e. a quotient that removes the powers of 2 and let the odd numbers be the representative of every costet
*coset
Then consider the set $\Bbb Z[\frac12]\setminus 3\Bbb Z[\frac12]/\sim$
This is just the odd numbers which are not multiples of 3
But as representatives of their cosets, they represent the dyadic rationals and even numbers too
Then $f(x)=3x+1$ or $f(x)=3x+2^{\nu_2(x)}$ is the Collatz function and that it converges to $<1>$ is the conjecture.
Then the nice properties of the relation are:
a) it's orthogonal to $f$
b) combined with the two smallest representatives it generates all the immediate predecessors of any given number
c) If we were to include the multiples of 3, it preserves the "has a predecessor" property - i.e. it is isometric in $\lvert\cdot\rvert_3$
d) it commutes with 2x
i.e. x\sim y\iff 2x\sim 2y
Sorry I have committed an unkind act by reusing $\sim$ for a different purpose
e) It repeats the form 3x+1
@RobertFrost I don't see what knowing the immediate predecessors should help. -- What MIGHT help is finding a suitable relation x \sim f(x) where there is no upper bound on the distance of x and f(x) in the Collatz graph.
But that is MUCH harder.
18:09
@StefanKohl Yes, that would of course prove the conjecture. The question was this really... In my equtions here there is transitivity of the relations 2x and 3x, and also the form (3x+1) replicates from the forward movement in the Collatz graph, into a lateral movement in the same graph. I was seeing scope this might be the transitivity you were looking for in your group action. But it seems you don't see how this could be the case?
@RobertFrost No, this would not "of course prove the conjecture". -- It would still depend very much on what the relation looks like, how it is given and what you can tell about it. -- It may very well happen that you are still at a dead end then.
@StefanKohl On another note, you say "I don't see what knowing the immediate predecessors should help". This is my belief: The Collatz graph can be represented as a Torsion group in which $3x+1$ moves from the set of element of orders (groupwise) n to those of order n-1. I.e. towards the group identity. Then $f$ as expressed above is an epimorphism on a non-hopfian torsion group. If this is correct then seeing immediate predecessors at least permits construction of the graph of the group.
This structure is isomorphic to the Prufer p-groups but has order $\omega$
instead of p
@StefanKohl Anyway, thank-you for chatting. I'm setting off home now but if you have anything more to say I will check back later
@RobertFrost Thank you also.

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