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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.

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    $\begingroup$ What do you mean by "exactly two thirds"? (There isn't an obvious way to make sense of a fraction of an infinite set.) $\endgroup$ – Noah Schweber Jun 5 '18 at 12:59
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    $\begingroup$ 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? $\endgroup$ – Servaes Jun 5 '18 at 13:33
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    $\begingroup$ 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. $\endgroup$ – user1729 Jun 5 '18 at 14:05
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    $\begingroup$ (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.) $\endgroup$ – user1729 Jun 5 '18 at 14:07
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    $\begingroup$ 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. $\endgroup$ – Servaes Jun 5 '18 at 14:30
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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.

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  • $\begingroup$ 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. $\endgroup$ – samerivertwice Jun 6 '18 at 20:38
  • $\begingroup$ Does this work: Let $G$ be a group satisfying $G=\{0,1,2\}\times F$ then two-thirds of $G$ are squares if $G^2=\{1,2\}\times F$ $\endgroup$ – samerivertwice Jun 7 '18 at 8:56
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    $\begingroup$ @RobertFrost Modulo an appropriate level of formality, yes your idea seems fine. $\endgroup$ – user1729 Jun 7 '18 at 9:34
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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.)

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    $\begingroup$ Why is $H$ a subgroup of $G$? Is $G$ assumed to be abelian? $\endgroup$ – Servaes Jun 5 '18 at 13:29
  • $\begingroup$ 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). $\endgroup$ – samerivertwice Jun 5 '18 at 14:12

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