# Does $abab=baba$ imply commutativity in a Group of uneven order?

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.)
@DougSpoonwood: Note that for every $g$ in a finite group $G$; $g^{|G|}=e_G$ so, $ab=(ab)e_G=(ab)(ab)^{2k-1}$. –  B. S. Jun 9 '12 at 3:10