The starting point of this question is the following:



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



Are infinite groups in which most elements have order $\leq 2$ commutative?
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.


