TITLE: Are infinite groups in which most elements have order $\leq 2$ commutative?
QUESTION [23 upvotes]: 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?

REPLY [5 votes]: 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.

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.

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.