Timeline for An approximate version of $g^2 = e$ for all $g \in G$, implies $G$ is Abelian

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Sep 29, 2023 at 16:28	comment	added	Z. A. K.		NB the "degree of satisfiability" bound for $x^2 = e$ is precisely the $\frac{3}{4} < \sqrt{\frac{5}{8}}$ taken in $D_8$. This has an easy proof without invoking Gustafson's result at all: when $x^2=e$ in such a group, then $xy=yx$ holds whenever $y^2=(xy)^2$. But if more than $\frac{3}{4}$ of the elements satisfy $g^2=e$, then both sides equal $e$ with probability over $\frac{1}{2}$. So the centralizer of $x$ is larger than half the group - by Lagrange it is the whole group, and $x\in Z(G)$. But then $Z(G)$ is more than $\frac{3}{4}$ of the group, so invoking Lagrange once again, $Z(G)=G$.
Aug 25, 2015 at 15:11	history	edited	Geoff Robinson	CC BY-SA 3.0	Simplified
Aug 24, 2015 at 16:11	history	edited	Geoff Robinson	CC BY-SA 3.0	corrected inaccuracy
Aug 24, 2015 at 13:27	comment	added	Joseph O'Rourke		And see also,"Why can't a nonabelian group be 75% abelian?," also answered by Geoff, where the $\frac{5}{8}$ is explained.
Aug 24, 2015 at 13:08	history	edited	Geoff Robinson	CC BY-SA 3.0	minor rewording
Aug 23, 2015 at 19:33	history	edited	Geoff Robinson	CC BY-SA 3.0	Mentioned connections to Theorems of Frobenius et al
Aug 23, 2015 at 18:39	history	answered	Geoff Robinson	CC BY-SA 3.0