TITLE: Prove that $G$ is abelian
QUESTION [18 upvotes]: Let $G$ be a group with the property that for any set of three distinct elements in $G$, say $x$, $y$, $z$, at least two of them will commute. Prove that $G$ is abelian.

I have no idea how to start this problem. How do I use the fact that two elements for any set of three elements of $G$ commute.

REPLY [3 votes]: Suppose there are $a,b$ that don't commute, and denote the centralizer of $x$ as $C(x)$. 
The hypothesis says that $G=C(a)\cup C(b)$. But a group can't be the union of two proper subgroups, so one of these, say $C(a)$, equals $G$. 
But then $a$ is central and commutes with $b$, a contradiction. Thus $G$ is abelian.