Prove that $G$ is abelian

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.

-
Do you know the lemma that a group is never the union of two proper subgroups? That makes this problem really simple (I used this strategy in my solution.) – rschwieb Dec 7 '15 at 18:34

Given $a,b\in G$, consider the three elements $a,b$, and $a^{-1}b^{-1}$. If $a$ and $b$ don’t commute, then $a^{-1}b^{-1}$ commutes with at least one of $a$ and $b$. Say it commutes with $b$: then $$aba^{-1}b^{-1}=aa^{-1}b^{-1}b=1_G\;,$$ and $a$ and $b$ commute. If instead $a^{-1}b^{-1}$ commutes with $a$, $$a^{-1}b^{-1}ab=aa^{-1}b^{-1}b=1_G\;,$$ and again $a$ commutes with $b$.

Added: As Steve D points out, $a^{-1}b^{-1}$ may be equal to one of $a$ and $b$, in which case I don’t actually have three elements of $G$. If $a^{-1}b^{-1}=a$, then $b=a^{-2}$, which certainly commutes with $a$, and the argument is similar if $a^{-1}b^{-1}=b$.

-
There is the (easy) case you have ignored: if $a^{-1}b^{-1}$ is equal to either $a$ or $b$. I think generally the solution is to use $ab$ as the third element, and note that commuting with $a$ is the same as commuting with $a^{-1}$. – user641 Apr 7 '12 at 10:15
@Steve: Thanks; fixed. – Brian M. Scott Apr 7 '12 at 10:20
You could also consider $a$, $ab$ and $ba$.If $a$ does not commute with $b$, then $a$, $ab$ and $ba$ are distinct. So $a$ commutes with either $ab$ or $ba$. Multiplying the relation with $a^{-1}$ (to the left or the right), you get that $a$ and $b$ commute. – Joel Cohen Apr 7 '12 at 12:11

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.

-