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Suppose $(G,\cdot)$ is a finite group of uneven order such that $abab=baba$ for any $a,b\in G$. Does this mean that $G$ is commutative?

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"uneven" is a long form of "odd"? :) – Mariano Suárez-Alvarez Jun 8 '12 at 22:26
"uneven"="non-even"="even $\cup$ infinite"?... – user1729 Jun 8 '12 at 22:49

Yes. Let $|G|=2k-1$ be the order of the group and $a,b\in G$. Then: $$ab=ab(ab)^{2k-1}=(ab)^{2k}=(abab)^k=(baba)^k=(ba)^{2k}=ba(ba)^{2k-1}=ba.$$

(Added: I should probably mention that here we use the following fact twice: if $G$ is a finite group of order $n$ and $a\in G$, then $a^n=e$, where $e$ is the identity element.)

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How do you have that ab=ab(ab)^{2k−1} and ba(ba)^{2k−1}=ba? – Doug Spoonwood Jun 9 '12 at 2:35
@DougSpoonwood: Note that for every $g$ in a finite group $G$; $g^{|G|}=e_G$ so, $ab=(ab)e_G=(ab)(ab)^{2k-1}$. – Babak S. Jun 9 '12 at 3:10
@DougSpoonwood: I am using the fact Babak mentions. The fact itself follows from elementary divisibility properties of natural numbers. It may also be seen as an easy consequence of Lagrange's theorem. You're quite right, I should have mentioned this from the beginning. – Dejan Govc Jun 9 '12 at 10:05

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