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Can a countably infinite torsion group exist in which the density of squares is double the density of non squares?

By $x\in G$ being a square I simply mean there exists $y\in G$ satisfying $y\cdot y=x$

Does the question have different answers if we assume the group to be finitely, or infinitely generated?

Let me define "double the density" more precisely.

There exists a numbering of $G$ such that for every set $S_x=\{y: y\circ y=x\}$, there is a least element $S_x(0)=\min{(S_x)}$ which can be numbered either $\equiv 0,1,$ or $2\pmod{3}$ and every element of $S_x$ which is numbered $\equiv n\pmod{3}$ has a unique right-successor in $S_x$ which is numbered $\equiv n+1\pmod{3}$.

Then every element which is numbered $\equiv1,2\pmod{3}$ is a square of infinitely many elements and every element numbered $\equiv0\pmod{3}$ is a square of none.

Robert Frost
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    What do you mean by "exactly two thirds"? (There isn't an obvious way to make sense of a fraction of an infinite set.) – Noah Schweber Jun 05 '18 at 12:59
  • @NoahSchweber :( I did wonder about that. The scenario is that the group's set is the odd integers and those $\equiv 1,2\pmod{3}$ have a square root while those $\equiv 0$ don't. What isn't obvious from that statement is that the $\frac{2}{3}:\frac{1}{3}$ rule applies perfectly among all numbers irrespective of the exponent of their square. i.e. for every $x$ for which there exists some $y$ satisfying $y\cdot y=x$, there are infinitely many $y$ and they obey the two-thirds rule too. You could think of them as successor segments in the usual construction of the Cantor set. – Robert Frost Jun 05 '18 at 13:11
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    This doesn't clarify what you mean by "two thirds". Also, under what operation do odd integers congruent to 1 or 2 mod 3 have a square root while those divisible by 3 don't? – Servaes Jun 05 '18 at 13:33
  • @Servaes I don't really want to attempt to draw somebody into tackling the Collatz conjecture, rather learn what group structures are possible in principle. But the answer to your question is that the operation would be $f:\Bbb N\setminus 2\Bbb N \to \Bbb N\setminus 2\Bbb N, f(x)=(3x+1)\lvert 3x+1\rvert_2$ and $G=(\Bbb N\setminus 2\Bbb N,\circ)$ where $x\circ x=f(x)$. The question really asks whether a) a torsion group structure is possible & b) whether extension to some domain greater than $\Bbb N\setminus 2\Bbb N$ is necessary. But this doesn't define the group as $\circ$ isn't well-defined. – Robert Frost Jun 05 '18 at 13:51
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    As a quick comment: if $G$ is finitely generated then you can discuss properties of elements of length $\leq n$, and then let $n\rightarrow\infty$. I think this idea is due to Gromove; it is certainly related to his "random group" idea. – user1729 Jun 05 '18 at 14:05
  • @NoahSchweber I've added a precise definition of "two thirds of elements having a square root" to the question. – Robert Frost Jun 05 '18 at 14:06
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    (As an aside, I do not understand the addition of the "collatz" tag. There is no obvious connection, even if this is the OPs motivation.) – user1729 Jun 05 '18 at 14:07
  • @user1729 thanks. I believe it's not finitely generated only because of various growth heuristics of the Collatz conjecture, althought I don't know. Re Collatz, the comment to Servaes precisely defining the conjecture above perhaps explains this. – Robert Frost Jun 05 '18 at 14:10
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    It is still completely unclear to me. Your definition seems to suppose that every $x\in G$ has infinitely many square roots? And the condition that $S_x$ permits an enumeration in which every third element is not a square is equivalent to $S_x$ containing both infinitely many squares and non-squares. – Servaes Jun 05 '18 at 14:30
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    Also, what does $|3x+1|_2$ mean in the definition of $f(x)$? – Servaes Jun 05 '18 at 14:31
  • @Servaes I didn't intend to imply that every $x$ has infinitely many square roots and that's not the case - I've clarified the question. Yes, every $S_x$ contains both infinitely many squares and non-squares (although the numbering of them implies the measure of the non-squares is half the measure of the squares in the set). $|3x+1|_2$ is the 2-adic absolute value. – Robert Frost Jun 05 '18 at 14:36
  • This still seems to imply that if $x\in G$ has a square root, then it has infinitely many. This is a strong condition on $G$. And are you looking for an enumeration of $G$ that has this property for every set $S_x$, or for separate enumerations of every set $S_x$? – Servaes Jun 05 '18 at 14:38
  • @Servaes yes, that's precisely the condition. – Robert Frost Jun 05 '18 at 14:40
  • @Servaes re your question about the enumeration I'm looking for, if my conjecture is true then an enumeration with this property exists for both the individual $S_x$ and another enumeration with the same properties exists for $\cup_x S_x=G$ although the ordering $0,1,2$ is the reverse i.e. $2,1,0$ for $G$. – Robert Frost Jun 05 '18 at 15:15
  • @Servaes just a little comment in relation to what you said about infinitely many square roots being a strong condition. If we take the direct sum of infinitely many prufer 2-groups, this torsion group has the property of every element having infinitely many square roots. But it lacks the one third of elements have none, of course. – Robert Frost Jun 05 '18 at 16:27

2 Answers2

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I voted to close this question as it is unclear. I have now changed my mind, and instead feel that it would be more beneficial to write an answer explaining:

  1. why the question is unclear, and

  2. how to make the underlying issue formal.

So, the question is unclear because the description of "double the density" does not make sense - a countable set can be "numbered" is basically any way. For example, it is "clear" that half of the integers are even...but we can number our integers by the following function: $$f(n)=\begin{cases} 3n/2 & n=2k\\ 3(n-1)/2+1&n=4k+1\\ 3(n-1)/2-1&n=4k-1 \end{cases}$$ Then, by the OPs logic, one third of the integers are even! This is silly, and why I voted to close the question as unclear.

Now, how should we make it formal that "half" of the integers are even? Well, one way is to look at the Cayley graph of $\mathbb{Z}=\langle 1\rangle$. Consider the closed ball of even radius $r=2k$ centered at $0$. Then $k+1$ of the elements in this ball are even numbers, so the proportion of even numbers is $(k+1)/(2k+1)$. So take the limit (use $k$ not $r$ as formula changes if $r$ is odd): $$\begin{align*} \lim_{k\rightarrow\infty}\frac{k+1}{2k+1} &=\lim_{k\rightarrow\infty}\frac{k+1}{2k+1}\times\frac{1/k}{1/k}\\ &=1+\lim_{k\rightarrow\infty}\frac{1+\frac1k}{2+\frac1k} \\ &=\frac12 \end{align*}$$

Okay, so how to do this in groups, rather than just in $\mathbb{Z}$? Well, you might be interested in the paper Y. Antolin, A. Martino, E. Ventura, Degree of Commutativity of Iinfinite Groups, Proc. Amer. Math. Soc. 145.2 (2017): 479-485 (link). They use the above idea to generalise an old result on finite groups. The "degree of commutativity" of a finite group $G$ is defined as:

$$\operatorname{dc}(G) =\frac{ |\{(u, v) \in G^2 : uv = vu\}|}{|G|^2}$$

roughly speaking, it is the proportion of commuting elements in your group. Note that $\operatorname{dc}(G)=1$ if and only if $G$ is abelian. The classical result is:

Theorem (Gustafson, 1973). Let $G$ be a finite group. If $\operatorname{dc}(G) > 5/8$, then $G$ is abelian.

The authors of the above paper used balls in Cayley graphs, as in my above example of $\mathbb{Z}$, to define the degree of commutativity of a group $G$ given by generating set $X$, denoted $\operatorname{dc}_X(G)$. They proved the following:

Theorem. Let $G$ be a finitely generated residually finite group of subexponential growth, and let $X$ be a finite generating set for $G$. Then:

  1. $\operatorname{dc}_X(G) > 0$ if and only if $G$ is virtually abelian;

  2. $\operatorname{dc}_X(G) > 5/8$ if and only if $G$ is abelian.

user1729
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  • Thank-you for this very interesting answer. I have been reflecting on more rudimentary attempts. One idea I was thinking of is whether the group could be a direct sum or product of Z3 or have Z3 as a subgroup. Then we might have an argument that the products of 0 have no root other than themselves. Can't really see how to make this simple approach hang together though. – Robert Frost Jun 06 '18 at 20:38
  • Does this work: Let $G$ be a group satisfying $G={0,1,2}\times F$ then two-thirds of $G$ are squares if $G^2={1,2}\times F$ – Robert Frost Jun 07 '18 at 08:56
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    @RobertFrost Modulo an appropriate level of formality, yes your idea seems fine. – user1729 Jun 07 '18 at 09:34
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As pointed out in the comments, two thirds of an infinite group is not necessarily well-defined. But you could reasonably say if $H$ is a subgroup of $G$ with finite index $q$, then the union of $p$ distinct cosets of $H$ makes up $\frac{p}{q}$rds of $G$.

In that sense, the answer is no. Let $H=\{y^2|y\in G\}$ - then $H$ is a subgroup of $G$ so it makes up exactly $\frac{1}{[G:H]}$rds of $G$. That's at most $\frac{1}{2}$ if $H$ isn't the whole thing; it can't be 2/3rds.

($\mathbb{Z}/3\mathbb{Z}$ isn't an example either. Additively, every element is a square root. Multiplicatively, there's 2 elements, one of which is a square root.)

anonymous
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    Why is $H$ a subgroup of $G$? Is $G$ assumed to be abelian? – Servaes Jun 05 '18 at 13:29
  • Thank-you. Is this compatible with the precise definition of "two-thirds of elements" now added to the question? The group is more likely abelian than not. Does your answer depend upon the group being abelian? I'm not clear just now that we can assume $H$ would be closed. In fact my intuition is that it isn't (but please don't delete this helpful answer). – Robert Frost Jun 05 '18 at 14:12