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Let $G$ be a group with the property that in every subset of 4 distinct elements, there exists at least a pair of commuting elements. Show that G is Abelian.

  • I have thought so far that if G isn't abelian then if x,y dont commute and given subset of index 3 then the subset $\{x,y,xy\} \implies xy = yx$. Can I find something similar for the 4-element case?
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  • $\begingroup$ $xy = yx \iff x(yx) = (yx)x \iff x(xy)=(xy)x\iff (yx)y=y(yx)\iff (xy)y = y(xy)$. $xy=yx\implies xyyx=yxxy$ and $xyyx=yxxy\implies...$ hmm, you may be right. I was assuming nick's calculations were good. $\endgroup$
    – fleablood
    Nov 9, 2017 at 22:59

1 Answer 1

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The theorem is incorrect. The quaternion group $Q_8$ is a counterexample, where $Q_8 = \{1, i, j, k, -1, -i, -j, -k\}$.

In this group, every element commutes with its negative. Also, $1$ and $-1$ commute with eveything. By the pidgeonhole principle, every 4-element subset of $Q_8$ either contains $1$, $-1$, or both an element and its negative. Hence, there is a commuting pair in every 4-element subset of $Q_8$.

But $Q_8$ is non-abelian, since $ij \ne ji$.

Thanks to @DanielFisher for his help.

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