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The starting point of this question is the following:

If $G$ is a group such that all elements have order at most $2$, then $G$ is commutative.

If $G$ is any group, let $G_{>2}$ denote the set of elements $g\in G$ such that $g^2 \neq 1_G$ where $1_G$ denotes the neutral element of the group.

Question. If $G$ is infinite, directly indecomposable, and $|G_{>2}|<|G|$, is $G$ necessarily commutative?

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    $\begingroup$ Assuming that $G$ is directly indecomposable makes this question more interesting. $\endgroup$
    – Denis T.
    Sep 18 '18 at 6:35
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    $\begingroup$ @ArturoMagidin I don't agree. If $H\neq 1$ has no element of order 2 and $P$ is a huge 2-elementary group, then in $H\times P$, the elements of order $\le 2$ are precisely elements of $P$, so elements not of order $\le 2$ form the complement $(H\times\{1\})\times P$ which is as large as $P$. So it's not relevant to discard direct products. $\endgroup$
    – YCor
    Sep 18 '18 at 7:13
  • $\begingroup$ It's not a direct answer to the question, but you might be interested in the following more quantitative theorem of Tointon. He proves that for finitely generated groups, if 'most' elements commute then the group is abelian. Clearly, for a suitable definition of 'most', if 'most' elements are of order 2 then 'most' elements commute. arxiv.org/abs/1707.05565 $\endgroup$
    – HJRW
    Sep 18 '18 at 8:14
  • $\begingroup$ @YCor: Urgh; you're of course right. Thank you. $\endgroup$ Sep 18 '18 at 16:58
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Let $G$ be an infinite group in which the set of elements of order $\neq 2$ has cardinal $<|G|$. Then $G$ is 2-elementary abelian.

Indeed, by contradiction, let $g\in G$ be of order $3\le d\le \infty$. So the conjugacy class of $g$ has cardinal $<|G|$, and hence the centralizer $C_g$ of $g$ has order $|G|$, and in turn, the set $C_g^{(2)}$ of elements of order 2 in $C_g$ has cardinal $|G|$. Then for every $h\in C_g^{(2)}$, the element $gh$ does not have order 2 (as $(gh)^2=g^2h^2=g^2\neq 1$). Since $h\mapsto gh$ is injective, this produces $|G|$ elements of order $\ge 3$, a contradiction.

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    $\begingroup$ Note that the same argument shows that in an infinite group $G$, if for some $k\ge 1$, the set of $g$ such that $g^k\neq 1$ has cardinal $<|G|$, then $g^k=1$ for all $g\in G$. $\endgroup$
    – YCor
    Sep 18 '18 at 8:55
  • $\begingroup$ It's a brilliant approach, thanks @YCor! $\endgroup$ Sep 18 '18 at 10:38
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    $\begingroup$ I'm curious; do you see a natural way to extend your proof to one about measures? I think there should be something interesting (and probably different from the conclusion I got). $\endgroup$
    – user44191
    Sep 19 '18 at 3:23
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As an alternative approach with the same solution:

Given $g\in G$, I will show that $g$ is of order $2$ (or $1$). From here on, a subset of $G$ is "large" if its complement is of lower cardinality than that of $G$.

Let $G^{(2)}$ be the set of elements of order $2$. Then $G^{(2)}z$ is large, so $G^{(2)} \cap G^{(2)}z$ is large. We can therefore choose $x$ such that $x, xg$ are both of order $2$.

Similarly, we can find $y$ such that $y, z:= yxg, xy, xz = xyxg$ are all of order $2$. Then by the usual proof, $x, y, z$ all pairwise commute and are of order $2$, so $xyz = xyyxg = xxg = g$ is of order $2$ (or $1$).

Interestingly, this approach should be easy to adapt to measure theory, leading to a conclusion of the form: Let $G$ be a group with a finite one-sided invariant finitely-additive measure (and assume the measure is normalized to $1$). Then if the set of elements of order $2$ is measurable with measure larger than $c$, the group is commutative and every element has order $2$. The limiting step in the above proof is the choice of $y$, and there are $4$ conditions on $y$, so $c = \frac{3}{4}$ should work. Considering the dihedral group of the square, in which $6$ out of $8$ elements are of order $2$ or less, $\frac{3}{4}$ should be optimal.

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  • $\begingroup$ Nice, thanks for this different angle! $\endgroup$ Sep 18 '18 at 10:37
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    $\begingroup$ Note that $xy$, resp $xyxg$, has order 2 iff $yx$, resp $yxgx$ has order 2. So this proof shows that, denoting by $G^{(2)}$ the set of elements of order $\le 2$, if $G$ is any group such that $G\neq G^{(2)}$ then there exist $g,h$ such that $G^{(2)}\cap G^{(2)}g\cap G^{(2)}h=\emptyset$. In particular, it shows the statement for $c=2/3$ (with 1-sided invariance of the measure, for which finite additivity is enough). $\endgroup$
    – YCor
    Sep 18 '18 at 18:16
  • $\begingroup$ @YCor I think you may have made an error somewhere; consider $D_8$ (or $D_8 \times C_2^\infty$). I now think that $\frac{3}{4}$ is optimal, since the limiting step is the choice of $y$, and $y$ must satisfy $4$ conditions. $\endgroup$
    – user44191
    Sep 19 '18 at 0:12
  • $\begingroup$ @YCor You're right, I didn't read the question properly. Thanks! $\endgroup$ Sep 19 '18 at 0:23
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    $\begingroup$ @user44191: you're right, I just didn't see that $y$ should also be of order 2, so there are indeed 4 elements. So the correct statement is the following: for any group such that $G\neq G^{(2)}$, there exist $g,h,k$ such that $G^{(2)}\cap G^{(2)}g\cap G^{(2)}h\cap G^{(2)}k=\emptyset$. So $c=3/4$ works, and, as you say, is optimal because of the example of $D_8$. $\endgroup$
    – YCor
    Sep 19 '18 at 0:29
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Here is a slight modification of YCor's solution, which is too long to describe in a comment. It is proved in the same way.

Claim. Any identity $w(x_1,\ldots,x_n)\approx 1$ which hold almost everywhere in an infinite group must hold everywhere.

Here, an $n$-ary identity $w\approx 1$ holds almost everywhere in infinite $G$ means the solution set $S\subseteq G^n$ of $w(x_1,\ldots,x_n) = 1$ satisfies $|G^n-S|<|G^n|=|G|$.

Step 1. If $w\approx 1$ holds almost everywhere, then so does $w(x,x,x,x,\ldots,x) \approx 1$, and this has the form $x^k=1$ for some $k$ (possibly $k=0$). As noted YCor's comment to his solution, this implies $x^k=1$ holds everywhere. Thus we may assume that $w(x,x,\ldots,x)\approx 1$ holds everywhere.

Step 2. If $w\approx 1$ did not hold everywhere, then there would exist a tuple $t=(g_1,\ldots,g_n)\in G^n$ that does not satisfy it. Each conjugate of $t$ fails $w\approx 1$, so the index of the centralizer of $t$ is small, forcing $|C_G(t)|=|G|$.

Step 3. For each $h\in C_G(t)$ we have $$ w(hg_1,hg_2,\ldots,hg_n) = w(h,h,\ldots,h) w(g_1,g_2,\ldots,g_n) = 1\cdot w(g_1,g_2,\ldots,g_n) \neq 1, $$ yielding $|G|$-many failures of $w\approx 1$, namely all tuples in $C_G(t)\cdot t$. This is too many failures of $w\approx 1$, thereby contradicting the existence of even one failure $t$ of $w\approx 1$.

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  • $\begingroup$ In Step 3 it's enough to say "almost everywhere" for $h$, so you do not really need my statement, your proof does all the job. $\endgroup$
    – YCor
    Sep 18 '18 at 18:48
  • $\begingroup$ @YCor: Let's not confuse whose proof it is! (yours) $\endgroup$ Sep 18 '18 at 19:30
  • $\begingroup$ Sure, but I just mean that your argument is self-contained, even if inspired by mine. $\endgroup$
    – YCor
    Sep 18 '18 at 19:31
  • $\begingroup$ I am not seeing the need for Step 1. Aren't there enough h's in the centralizer already? Gerhard "Maybe Ycor Means Precisely This" Paseman, 2018.09.18. $\endgroup$ Sep 18 '18 at 19:39
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    $\begingroup$ @GerhardPaseman Yes. Y "I precisely mean this" Cor, 2018.09.18. $\endgroup$
    – YCor
    Sep 18 '18 at 19:50
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Here is some thougths on infinite finitely generated groups.

Let $sq\colon G\to G$ be the square function mapping $g$ to $g^2$. Hence the set of elements of order at most $2$ is equal to $sq^{-1}(1)$. Then, as proved in other answers, if $G-sq^{-1}(1)$ is small, then $G$ is abelian.

In the specific case of finitely generated groups, more can be said. Observe that such groups are countable, and hence "small" equal finite in this context.

  1. If $G$ is finitely generated, then $sq(G)$ is infinite as soon as $G$ is infinite. In particular, if there exists $F$ a finite subset of $G$ such that $G-sq^{-1}(F)$ is finite, then $G$ is finite.

  2. If $G$ is finitely generated and there exists $F$ a finite subset of $G$ and $g\in G$ such that $G-\bigl(sq^{-1}(F)\cup gsq^{-1}(F)\bigr)$ is finite, then $G$ is virtually abelian.

In this setting, it is not possible to conclude that $G$ is abelian. Counter-examples include the infinite dihedral group (with $F=\{1\}$) and generalized dicyclic group $Dic(A,x)$ (with $F=\{x^2\}$ not containing $1$).

The proof of 1. consists of applying carefully Dicman’s Lemma, which says that if $N$ is a finite normal subset of $G$ consisting of torsion elements, then $\langle N\rangle$ is a finite normal subgroup. The proof of 2. is more convoluted and use arguments on random walks due to Tointon. See https://arxiv.org/abs/2010.06020 for the details and some related statements.

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